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Penguin library of physical 
sciences: Physics/Chemistry 


Gases, liquids and solids 

D. Tabor 









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Gases, Liquids and Solids r 

D. Tabor 

Penguin Books 

Penguin Books Ltd, Harmondsworth, 

Middlesex, England 

Penguin Books Inc., 7110 Ambassador Road, 

Baltimore, Md 21207, U.S.A. 

Penguin Books Australia Ltd, Ringwood, 

Victoria, Australia 

First published 1969 

Reprinted 1970 

Copyright © D. Tabor, 1969 

Made and printed in Great Britain by 
Bell & Bain Ltd, Glasgow 
Set in Monotype Times 

This book is sold subject to the condition that 
it shall not, by way of trade or otherwise, be lent, 
re-sold, hired out, or otherwise circulated without 
the publisher's prior consent in any form of 
binding or cover other than that in which it is 
published and without a similar condition 
including this condition being imposed on the 
subsequent purchaser 


Editorial Foreword ix 

Preface xi 

Chapter 1 Introduction 1 

Chapter 2 Atoms, molecules and the forces between them 3 

2-1 Atoms and molecules 3 

2-2 The forces between atoms and molecules 13 

Chapter 3 Temperature, heat and the laws of thermodynamics 23 

3-1 Temperature 2S 

3-2 Heat 24 

3-3 The laws of thermodynamics 25 

Chapter 4 Perfect gases - bulk properties and simple theory 33 

41 Bulk properties 33 

4-2 Elementary kinetic theory of ideal gases 41 

4-3 The ether theory of the ideal gas 43 

4-4 Some deductions from kinetic theory 44 

4-5 Transport phenomena 50 

4-6 Sound waves in a gas 60 

Chapter 5 Further theory of perfect gases 65 

51 A better kinetic theory 65 

5-2 Sedimentation 74 

5-3 Temperature variation of reaction rates 76 

5-4 Distribution of velocities in a perfect gas 77 

5-5 Thermal energy of molecules 85 

5-6 Macroscopic examples of equipartition of energy 92 

Chapter 6 Imperfect gases 97 

6-1 Deviations from perfect gas behaviour 97 

6-2 Kinetic theory of an imperfect gas: van der Waals equation 101 

v Contents 

6-3 Some properties of the critical point 113 
6-4 Law of corresponding states 115 
6-5 Expansion of gases 116 

Chapter 7 The solid state 122 

71 Types of solids 122 

7-2 Main types of bonding in crystalline solids 123 

73 Solid-liquid transitions 124 

7-4 Consequences of interatomic forces in solids 125 

Chapter 8 The elastic properties of solids 134 
8-1 Some basic elastic properties 134 

8-2 Propagation of longitudinal waves along an elastic bar 139 
8-3 Bulk moduli 140 
84 Elastic properties of rubber 148 

Chapter 9 The strength properties of solids 154 
9-1 Deformation properties 154 
9-2 Dislocations 161 
9-3 Brittle solids 165 
9-4 Conclusion 170 

Chapter 10 Thermal and electrical properties of solids 171 

101 Specific heat 171 

10-2 Thermal expansion: Gruneisen's law 179 

10-3 Thermal conductivity 184 

10-4 Electrical conductivity of metals 188 

Chapter 11 The liquid state 193 

1 The liquid as a modified gas 193 

2 The structure of liquids: the 'radial distribution function' 196 

3 The liquid as a modified solid 199 

4 The liquid state sui generis 199 

5 Ice and water 204 

6 General approach 206 

7 Latent heat of fusion 206 

8 Melting point 207 

9 Vapour pressure 208 

10 Surface tension 212 

11 Nucleation in condensation: the Wilson cloud chamber 220 

12 Superheating 221 

13 The energy for capillary rise 221 

Chapter 12 Liquids: their flow properties 223 
12*1 Flow in ideal liquids: Bernoulli's equation 223 
12*2 Flow in real liquids, viscosity 225 
12-3 Rigidity of liquids 233 
12-4 Non-Newtonian flow 234 

vi Contents 

Chapter 13 Dielectric properties of matter 235 

13-1 Basic dielectric relations 235 

13-2 Polarization of gases 237 

13-3 Polarization of polar molecules 241 

13-4 Optical dispersion and anomalous dispersion 246 

13-5 Dielectric properties of liquids and solids 248 

13-6 Summary 252 

Chapter 14 Magnetic properties of matter 254 
14-1 The magnetic equations 254 
14-2 Diamagnetism: Langevin's treatment 258 
14*3 Paramagnetism: the Langevin function 261 
14-4 Ferromagnetism 264 

14-5 Quantum treatment of magnetic properties 268 
14-6 Ferromagnetic domains 271 
14-7 Magnetic hysteresis 271 
14-8 Summary: comparison of dielectric and magnetic properties 273 

Appendix A c.g.s. and MKS units 276 

Appendix B Values of some physical constants 278 

Index 281 

vii Contents 

Editorial Foreword 

For many years, now, the teaching of physics at the first-degree level has 
posed a problem of organization and selection of material of ever-increasing 
difficulty. From the teacher's point of view, to pay scant attention to the 
groundwork is patently to court disaster; from the student's, to be denied the 
excitement of a journey to the frontiers of knowledge is to be denied his birth- 
right. The remedy is not easy to come by. Certainly, the physics section of the 
Penguin Library of Physical Sciences does not claim to provide any ready-made 
solution of the problem. What it is designed to do, instead, is to bring together 
a small library of compact texts, written by teachers of wide experience, 
around which undergraduate courses of a 'modern', even of an adventurous, 
character may be built. 

The texts are organized generally at three levels of treatment, corresponding 
to the three years of an honours curriculum, but there is nothing sacrosanct 
in this classification. Very probably, most teachers will regard all the first-year 
topics as obligatory in any course, but, in respect of the others, many patterns 
of interweaving may commend themselves, and prove equally valid in practice. 
The list of projected third-year titles is necessarily the longest of the three, 
and the invitation to discriminating choice is wider, but even here care has 
been taken to avoid, as far as possible, the post-graduate monograph. The 
series as a whole (some five first-year, six second-year and fourteen third-year 
titles) is directed primarily to the undergraduate; it is designed to help the 
teacher to resist the temptation to overload his course, either with the out- 
moded legacies of the nineteenth century, or with the more speculative 
digressions of the twentieth. It is expository, only: it does not attempt to 
provide either student or teacher with exercises for his tutorial classes, or with 
mass-produced questions for examinations. Important as this provision may 
be, responsibility for it must surely lie ultimately with the teacher, for he 
alone knows the precise needs of his students - as they change from year to 

Within the broad framework of the series, individual authors have rightly 
regarded themselves as free to adopt a personal approach to the choice and 
presentation of subject matter. To impose a rigid conformity on a writer is 
to dull the impact of the written word. This general licence has been extended 
even to the matter of units. There is much to be said, in theory, in favour of 
a single system of units of measurement - and it has not been overlooked 
that national policy in advanced countries is moving rapidly towards 

ix Editorial Foreword 

uniformity under the Systeme International (S.I. units) - but fluency in the use 
of many systems is not to be despised: indeed, its acquisition may further, 
rather than retard, the physicist's education. 

A general editor's foreword, almost by definition, is first written when the 
series for which he is responsible is more nearly complete in his imagination 
(or the publisher's) than as a row of books on his bookshelf. As these words 
are penned, that is the nature of the relevant situation: hope has inspired the 
present tense, in what has just been written, when the future would have been 
the more realistic. Optimism is the one attitude that a general editor must 
never disown! 

N. Feather 
January 1968 

x Editorial Foreword 


The subdivision of physics into mechanics; heat; light and sound; magnetism 
and electricity; properties of matter; and kinetic theory, goes back to the early 
days of classical physics. During the last twenty or thirty years there has been 
a discernible trend towards a regrouping of subjects, partly to allow for new 
knowledge and partly to allow for new methods of approach. Few of the areas 
have received such an impulse as those which, in classical days, were grouped 
together as kinetic theory and properties of matter. The new approach is to 
emphasize the atomic structure of matter and to show that by assuming the 
existence of attractive and repulsive forces between atoms and molecules and 
the presence of thermal energy, it is possible to explain nearly all the bulk 
properties of gases, liquids and solids in terms of relatively simple models. 

The present book, which attempts to do this, is based on the lecture 
course given to first year students at Cambridge. Its treatment is relatively 
simple and contains very little quantum physics or wave mechanics. It 
represents an attempt to bridge the gap between sixth-form physics and 
physical chemistry, and the more advanced courses which follow in later 
years of specialization. If it has any merit at all this is largely due to the 
devotion of many generations of Cavendish teachers who have hammered out 
the type of treatment and approach I have given here. I am particularly 
indebted to Dr T. E. Faber who first explained the Cavendish syllabus to 
me, and to Dr D. Shoenberg who has run a parallel course of lectures with 
me for the last few years and who has, on several occasions, saved me from 
perpetuating mistaken ideas. I also wish to thank the members of the 
Cavendish who have made comments on various parts of the manuscript, 
and Miss Shirley V. King of Birkbeck College for her comments on the first 
part of Chapter 1 1. Any mistakes, however, either of fact or of opinion are 
my own and I accept responsibility for them. 

Dr Leo Baeck, in his spirited critique of romantic religion, referred to 
orthodoxy in the following terms: 'that bent of mind which would cultivate 
respect for the answer but, in the process, often loses respect for the 
problem.' A text book such as this must naturally pay great respect to the 
answer. The student quite rightly expects to find the correct answer crisply 
and clearly given. This I have attempted to do. But it sometimes happens in 
physics, as in other walks of life, that the simple crisp answer is not the 
whole truth. Where it seems profitable I have discussed this as openly as 
possible. I hope it does not confuse the student, for this is not my intention. 

xi Preface 

I recognize that, so long as a student's ability is judged by examinations, 
both student and examiner will persist in cultivating great respect for the 
answer. On the other hand, the progress of physics in the long run, may well 
depend far more on our according greater respect to the problem. 

xii Preface 

Chapter 1 

The three states of matter are the result of a competition between thermal 
energy and intermolecular forces. It is this struggle which determines whether 
a given substance, under given conditions, is a gas, a liquid or a solid. Many 
textbooks which describe the bulk properties of matter tend to lose sight 
of this and to treat gases, liquids and solids as though they are quite unrelated 
materials. Indeed gases become an exercise in kinetic theory, solids an 
example of the laws of elasticity, and liquids a type of material which shows 
viscosity and surface tension. In the present book we attempt to show that 
many of the bulk properties of gases, liquids and solids can be explained 
in terms of intermolecular forces; in this way the three states are linked by a 
common property. We shall not give a detailed account of the origin of 
molecular forces but, assuming they exist, we shall show the part they play 
in determining the properties of matter. 

Chapter 2 is a very simple account of the nature of atoms and molecules 
and the forces between them. The third chapter deals with temperature and 
the concept of heat. These are the introductory chapters. 

The next three chapters describe the properties of gases. This is a well- 
worn area where the molecular approach has long been used to describe 
the bulk behaviour. Chapter 4 gives the simple kinetic theory of idea) gases, 
while Chapter 5 deals with a more sophisticated treatment which includes a 
simplified approach to the Boltzmann distribution, and a discussion of the 
equipartition of energy. Chapter 6 deals with imperfect gases where perfect 
gas behaviour is modified to allow for the finite size of the molecules and for 
the forces between them. 

We turn at once from gases to solids. In gases the molecules are virtually 
free, in solids they are bound to particular sites and their main thermal 
exercise consists in vibrating about their equilibrium positions. However, 
the molecular forces that cause the real gas to deviate from ideal gas 
behaviour are the same forces that are responsible for the existence of the 
solid state. In Chapter 7 we show that these forces can explain the heat of 
sublimation of solids, their thermal expansion, their elastic properties and the 
existence of the surface energy of solids. Some of these themes are analysed 
in greater detail in subsequent chapters. 

In Chapter 8 the elastic properties of solids are described in terms of 
intermolecular forces. This is applied quantitatively to ionic and van der 
Waals solids and empirically to metals. An account is also given of the 

1 Introduction 

elasticity of rubber. Chapter 9 deals with the plastic and brittle properties of 
solids and shows that these may be understood in molecular or atomic 
terms. For both types of strength property it is shown that the real strength 
is generally much less than the theoretical because of the presence of imper- 
fections. Chapter 10 describes the thermal and electrical properties of solids. 
In this chapter we give a brief account of the specific heats of solids in terms 
of atomic vibrations (the Einstein and Debye theories); we also show how 
thermal expansion arises as the amplitude of the vibration increases with 
increasing temperature. Finally we describe thermal and electrical conducti- 
vity; with metals these are both attributed to the transport of energy by the 
free electrons. The classical theory which treats the electrons as a gas gives 
the right answers for the wrong reason and it is shown how the theory must 
be modified to allow for the discrete energy-levels of the electrons. 

The next two chapters deal with the liquid state, the Cinderella of modern 
physics. The main difficulty here is that, although in some ways the inter- 
molecular forces dominate (a liquid occupies a specified volume unlike a gas), 
in other ways thermal motion dominates (a liquid shows mobility, unlike a 
solid). The molecules are both bound and free so that over small regions 
they appear to be highly ordered as in a solid whilst over large regions such 
order does not exist. These features and some recent descriptive ideas 
(particularly those of Bernal and of Alder) are given in Chapter 1 1 together 
with a molecular theory of vapour pressure and surface tension. In Chapter 12 
we discuss the flow of liquids and using the simplest form of Eyring's theory, 
show that viscosity may be understood in terms of intermolecular forces. 
Consequently the connection between surface tension and viscosity is found. 

The last two chapters deal with the dielectric and magnetic properties of 
matter. It is shown that these properties arise from the polarization of atoms 
and molecules produced by the applied electrostatic or magnetic field. The 
dielectric behaviour, described in Chapter 13, receives a very satisfactory 
explanation in terms of simple atomic and molecular models. In addition 
the treatment provides a direct correlation between dielectric properties, 
optical properties and van der Waals forces. The explanation of magnetic 
properties is more difficult since quantum effects are of major importance. 
The classical explanation of diamagnetism which we reproduce in Chapter 
14 yields the right answer but, at a more fundamental level, is quite wrong. 
Paramagnetism is more easily explained and, assuming the existence of 
strong interactions, a satisfactory and satisfying model for ferromagnetism 
is found. 

Some of the chapters lend themselves to brief summaries and where this 
has been possible they have been introduced at or near the end of the relevant 
chapter. A final word on units: the system adopted throughout this book is 
the c.g.s. system, but conversion tables are given at the back of the book. 

2 Gases, liquids and solids 

Chapter 2 

Atoms, molecules and the 
forces between them 

2-1 Atoms and molecules 

•1 *1 The evidence for atoms and molecules 

Matter is not a continuum of uniform density, but consists of discrete 
particles, or if one wishes to be more up to date, of localized regions of very 
high density separated by regions of almost zero density. The particulate 
nature of matter has been known to us since the time of the Greeks and the 
idea of atoms (units which could not be further cut or divided) is generally 
attributed to Democritus (c. 460-370 B.C.). Many of his concepts have a 
surprisingly modern ring and constitute a tribute to the immense power, as 
well as the originality, of the Greek approach to logical inference and 
abstract reasoning. Much of his work was, indeed, the result of thought rather 
than of direct experiment. One must not, however, read too much modern 
science into these early ideas. For example in a passage quoted by 
Theophrastus, De Sensu, 61-62, Democritus states, 'Hard is what is dense, 

and soft what is rare Hard and soft as well as heavy and light are 

differentiated by the position and arrangement of the voids. Therefore iron 
is harder and lead heavier.' But it would be wrong from this to attribute to 
Democritus a knowledge of dislocations and point defects. 

The first real attempt to get to grips with the basic 'atoms' of matter had 
to wait until more quantitative measurements and generalizations had been 
made. The law of multiple proportions, largely due to the work of J. Dalton 
(1766-1844) was one of these. It stated that if two elementary substances 
combined chemically to form more than one compound the weights of one 
which combine with a fixed weight of the other are in a simple ratio to one 
another. For example nitrogen and oxygen combine to form nitrous oxide, 
nitric oxide, nitrogen sesquioxide and nitrogen dioxide; three of these oxides 
were, in fact, amongst the compounds which Dalton first studied in the course 
of his investigation. In these compounds 14 grams of nitrogen combine 
respectively with 8, 1 6, 24 and 32 grams of oxygen : i.e. the ratio is 1 : 2 : 3 : 4. 
This fits in naturally with the concept of unit masses which can combine in 
simple multiples with one another. 

We consider a simpler case, the reaction of hydrogen and oxygen to form 
water. Quantitatively we find 1 gram of hydrogen combines with 8 grams of 
oxygen to form 9 grams of water. According to Dalton's atomic theory 

3 Atoms, molecules and the forces between them 

(Dalton 1808) we should write 

1 atom hydrogen + 1 atom oxygen -*■ 1 atom water 
or H + O ->• HO. 

When these reactions are studied in the gaseous state and the volumes are 
measured at a standard temperature and pressure, the equation becomes 

2 volumes hydrogen + 1 volume oxygen -»■ 2 volumes water vapour. 

The next step in the argument depends on the hypothesis put forward by 
A. Avogadro in 1811 that at a specified temperature and pressure a given 
volume of any gas contains the same number of unit masses. 
Our previous equation then becomes 

2 unit masses of hydrogen + 1 unit mass of oxygen — *■ 
2 unit masses of water. 

This statement makes sense only if the 'unit mass' of oxygen is itself divisible 
into two equal parts each of which goes into one of the unit masses of water. 
Arguments along these lines lead finally to chemical equations of the type 

2H 2 + 2 -- 2H 2 0, 

where H 2 , the smallest quantity of hydrogen which can exist in the free state, 
is called the molecule whilst H, the smallest quantity of hydrogen to enter 
into chemical combination, is called the atom. Avogadro's 'unit masses' are 
therefore molecules. 

2-1 -2 Avogadro's law 

This is still called a hypothesis but deserves to be promoted to the status 
of a law. It states that at a specific temperature and pressure a fixed volume 
of any gas or vapour contains the same number of free unit masses, i.e. 
molecules. If we re-define the atom of hydrogen as 1 unit mass then the 
molecule of hydrogen consists of 2 unit masses. These quantities in grams are 
the gram-atom and gram-molecule respectively (a gram-molecule of any gas 
at a pressure of 760 mm. Hg and a temperature of 0°C. fills approx. 
22,400 c.c.). The number of molecules in a gram-molecule of any substance 
is called Avogadro's number N and has the value of 6-06 x 10 23 . For 
example N molecules of hydrogen weigh 2 g., of oxygen 32 g., of gold 
197 g., of iodine 254 g. ; these constitute the gram-molecular weights of these 

2* 1 -3 First experimental deduction of the size of molecules 

Pioneer work by Fraulein Pockels in 1891 showed that certain organic 
molecules with polar end groups such as oleic acid which are insoluble in 
water in bulk will spread over the surface of clean water. Lord Rayleigh 

4 Gases, liquids and solids 

experimented further with this and showed in 1899 that if the amount of acid 
added to the water surface is less than a certain amount there is practically 
no reduction in surface tension of the water. If, however, the amount added 
exceeds a certain critical value the surface tension falls rapidly. This is 
shown schematically in figure 1. Rayleigh assumed that this change occurred 






amount of surface active material added 

Figure 1 . Schematic diagrams showing the surface tension of water as small 
quantities of surface active material are added. The surface tension remains fairly 
high until a critical amount corresponding approximately to a complete mono- 
layer of surfactant has been added ; it then falls rapidly to a constant lower value. 

when there was a continuous monomolecular layer of fatty acid on the 
surface. He was thus able to determine the weight of acid per sq. cm. of 
surface at which the monolayer was formed. Assuming the monolayer to have 
the same density as the acid in bulk the thickness of the monolayer was 
calculated and found to be about 10" 7 cm., i.e. 10 A. This was the first direct 
indication that atomic and molecular dimensions are of the order of 

2-1-4 Determination of Avogadro's Number N 

We shall mention three very different ways of determining N . They are all of 
historic interest but only the last two provide reasonably accurate values. 

1-4-1 Radioactive disintegration. Radium disintegrates to give off a particles and 
leave radon gas. Radon disintegrates to give off a particles and leave behind 
radium A. Further stages in the disintegration are not, in the present context, 
of great importance. The a particles emitted can be counted by allowing the 
particles emitted within a small known solid angle to strike a fluorescent 

5 Atoms, molecules and the forces between them 

screen. Each particle may be counted since it produces an individual scin- 
tillation on the screen. From the size of the solid angle chosen the total 
emission over a solid angle of 4n can be deduced. Lord Rutherford in 1909 
showed that the a particle is a He atom which has lost its two orbital 
electrons (see below). These electrons are picked up by the particles so that 
each a, particle becomes a helium atom, and since helium is a monatomic gas 
each a particle thus becomes a gas atom. The volume of the helium gas 
produced in this way can be measured. By extrapolation Rutherford found 
that, in one year, one gram of radium would emit a total of 1 1 6 x 10 7 
particles which would turn into 0-043 c.c. of helium gas at s.t.p. 

Consequently 0-043 c.c. He contains 1 1 -6 x 10 7 atoms. 
Therefore 22,400 c.c. He contain 6 x 10" atoms. 

This derivation is of interest since it showed for the first time that atoms 
are individual particles. 

2-1-4-2 Electrolysis. If a quantity of electricity known as a faraday is used to electro- 
lyse a dilute acid, exactly 1 gram of hydrogen is liberated. Since the faraday is 
equal to 96,500 coulombs, this means that 96,500 coulombs are involved in 
converting N ions of hydrogen into N atoms of hydrogen gas. The charge 
on the hydrogen ion is due to the loss of a single electron, hence 

96,500 coulomb = charge of N electrons = N e. 

The next step is to determine e. This was carried through successfully by 
R. Millikan in 1909. He produced very fine droplets of oil between the plates 
of a parallel plate condenser, the plates being set in a horizontal plane. The 
particles gradually fall because of gravity. If, however, the air surrounding 
them is subjected to ionizing radiation the particles will pick up charge of 
say q. It then becomes possible to apply an electrostatic field E which is able 
just to hold the drop in equilibrium in a stationary position. In this state 

Eq = mg. 

Therefore as m, g and E were known, q was calculated and found always 
to be an integral number of some fundamental unit of charge. This was 
assumed to be the charge e of the electron. The current value of e found in 
this way is very nearly 

Hence N = 96,500 , Q = 6-03 x 10". 
l-6xl0" 19 

2-1-4-3 X-rays. X-rays are electromagnetic waves of short wavelength (about 10" 8 
cm.). M. von Laue was the first to suggest, in 1912, that since the spacing 
of crystal planes in a simple inorganic crystal is of this order, it ought to be 
possible to use the crystal as a diffraction grating for X-rays. The waves 

6 Gases, liquids and solids 

length X of the X-rays may be determined accurately by diffracting them with 
a conventional ruled grating, at glancing incidence. The X-rays may then be 
used to find the lattice spacing in the crystal. If, for example, a collimated 
beam of X-rays strikes a crystal plane at an angle 0, , it is first refracted through 
an angle 6 X , but since the refractive index for X-rays is almost unity one may 
take 0, = d 2 = 6. The condition for reinforcement is then that 2d sin 
= nk, where d is the distance between scattering planes and n is an integer. 
By observing the angles for which strong diffracted beams are obtained, the 
various scattering planes and their separation may be found. In this way 
the crystal structure and the lattice spacings may be deduced. 

As an example consider the sodium chloride crystal. This has a cubic 
structure with sodium and chloride ions at alternate sites (see figure 2). 

Figure 2. Structure of sodium chloride crystal. The sodium and chloride ions are 
arranged alternately on a cubic lattice, the separation a between ions „ being 
2-818 A. The sodium fluoride structure is identical except that a = 2-31 A. The 
'unit cell' size is 2a. 

The distance a between each site is found by X-ray analysis to be 2 818 A. 
Consider one of the unit cuoes of side a. Each of these cubes has 4 sodium 
and 4 chloride ions, i.e. 4 NaCI molecules. But each corner of the cube is 
shared by eight contiguous cubes. Consequently each volume a 3 contains 
one eighth of four molecules of NaCI = i molecule of NaCI. If M = mole- 
cular weight, p = density, the volume occupied by the half molecule is 

1W1 „ 

— — - . Consequently 

2N p 

, 1M1 

a 3 = . 

2 N„p 

Inserting M = 58-5, p = 2-17, a = 2-818 A., 

N u = 602x10". 

7 Atoms, molecules and the forces between them 

2-1 -5 Charge, mass and configuration of atoms 

1. An atom consists of a minute nucleus, which is positively charged and 
contains the main mass of the atom. 

2. The nucleus consists of protons and neutrons. These are of equal mass 
but each proton carries a positive charge whilst the neutron carries no charge. 
The resultant positive charge (referred to the proton as unit charge) is called 
the atomic number Z, that is Z is equal to the number of protons in the 

3. As a crude approximation, valid especially for smaller atoms of atomic 
number less than 20, there are about equal numbers of protons and neutrons 
so that the mass of the nucleus, referred to the proton as unit mass, is about 
twice the atomic number. This is known as the mass number and in practice 
is the integer nearest to the mass of the atom, referred to the carbon- 12 atom 
as 12-000. In the older literature the mass number is sometimes loosely 
referred to as the atomic weight. 

4. The nucleus is surrounded by electrons. These have a mass only x*tt 
of the mass of the H atom (i.e. of a proton or neutron) and they have a unit 
negative charge. 

5. Since the atom is electrically neutral the number of orbital electrons 
is equal to the atomic number. 

6. The chemical properties of the atom are determined by the configuration 
of the electrons, especially of the outermost ones. 

7. There are certain stable arrangements, the electrons falling into well 
defined energy levels or shells, defined by quantum conditions. Each energy 
level can hold no more than two electrons and these must have opposite 
spins (see 8 below). This is known as the Pauli Exclusion Principle. 

8. The lowest energy shell, with principal quantum number 1, is the K 
shell. This ground-state is called the \s orbital. Here the electron is distributed 
with spherical symmetry about the nucleus and has zero angular momentum. 
However, the electrons can have a spin quantum number which can be de- 
scribed as being either parallel or antiparallel to some specified direction, 
for example, f or \. If the K shell contains 1 electron as in hydrogen it has a 
single spin; if it contains two electrons as in helium the spins are paired in 
opposite directions and the K shell is now complete. The atom is thus inert. 

9. The next higher energy shell has principal quantum number 2 and is 
known as the L shell. The lowest level is called the 2s state (again spherically 
symmetrical) and this can hold 2 electrons of opposite spin. There is a higher 
energy orbital known as the 2p. This has angular momentum, is dumb-bell 
in shape, and there are three directions available. Each can hold 2 electrons 
of paired spins so that altogether this state can hold 6 electrons. If only two 
or three electrons have to go into the 2p state the energy is lowest if they 
each have parallel spins (this is part of a more general principle, due to F. 
Hund, which says that when several levels within a given state are available 
the lowest energy corresponds to parallel spins): they can do this by each 

8 Gases, liquids and solids 

going into the three available orbitals. These electrons can readily play a 
part in chemical bonding and other processes. A fourth electron will have to 
pair its spin with one of the existing three. The remaining two unpaired 
electrons are then readily available for chemical reaction. When the Ip 
state is filled with its six electrons the atom - neon - is chemically non- 
reactive (but see below). 

10. The next higher level has principal quantum number 3 and is the M 
shell. This can hold 2 electrons in the 3s orbital, 6 in the 3/> and 10 in the 3d. 
(Again the d orbitals are directed and have angular momentum.) The next 
higher energy level is the N shell which contains 2 electrons in the 4s, 6 in the 
4p, 10 in the 4d and 14 in the 4/ states. However, it happens that the lowest 
state in the N shell (4s) generally has lower energy than the 3d states in the 
M shell: electrons will therefore go into the 4s state before they start filling 
the 3d orbitals. The arrangement of shells and stable sub-groups are sum- 
marized in Table 1. 

Table 1 Shells and stable sub-groups 




of electrons 

Number of electrons 


in stable sub-groups 












2, 6, 10 





As mentioned above if an atom contains a complete shell or sub-group it 
is relatively stable and may even be inert. In Table 2 we construct the first 
part of the periodic table based on the shell-model described above. 

We see that the outermost electrons, especially those with unpaired spins, 
determine the chemical properties of the atoms. As a simple approximation 
we may say that the valency reflects the way in which the atom loses, gains 
or shares electrons in an attempt to form stable shells or sub-shells. When 
the atom already contains such a configuration (He, Ne, and Ar above) 
it is already stable and is chemically inert. This needs some qualification 
since in recent years it has been shown that the larger inert atoms such as 
xenon can, in fact, be ionized by the presence of a powerful electronegative 
ion such as the fluoride ion. As a result it is possible to form a stable com- 
pound of xenon hexafluoride. 

9 Atoms, molecules and the forces between them 

Table 2 The beginning of the periodic table of elements 











Electronic states Is 






4 s 4/> 4d 4/ 

At. No. Atom 


































t| inert 

filled f 


filled f 

t. t 
t. t. t 

u. t, t 
U, H. t 

ti> ti. U inert 
filled t 


filled f 

H. t|. t| inert 
filled empty f 

empty t| 

t U 

filled f 






2*1 -6 Size of atoms 

The simplest way of forming some quantitative idea of the size of atoms 
is to measure the distance between atoms in the solid state. The result, of 
course, depends on the crystal structure. Some typical results are given in 
Tables 3 and 4. 

10 Gases, liquids and solids 

Table 3 Atomic diameters for vertical columns in periodic table 


Electron shells 
Z K L 







































The outermost configuration is unchanged. As we fit in additional shells or 
subshells there is an increase in d. (The inner orbitals contract because of the 
increase in nuclear charge.) 

Table 4 Atomic diameters for horizontal columns in periodic table 


Electron shells 
Z K 
































































Note that Ir (Z = 77) has smaller diameter than Al (Z = 13). d Qr) = 2-7 A., 
«/(Al) = 2-8 A. 

Here the main process is one of filling up an unfilled shell whilst the 
outer configuration scarcely changes. The increasing nuclear charge causes 
shrinkage in the atomic diameter. 

11 Atoms, molecules and the forces between them 

2'1'7 Size of ions 

Some results are given in Table 5 for monovalent atoms which either lose an 
electron to become a positive ion, or gain an electron to become a negative 
ion. It is seen that the effect of removing one electron from the atom is to 
reduce d by 1 -8 to 2 A. ; the effect of adding an electron is to increase d by 
about the same amount. 

Table 5 Effect of ionization on size 

(size obtained from spacing in appropriate compounds) 























Diameter dk. 


Atom Ion 


1-3 2-7 


2 3-6 


2-3 3-9 


2-7 4-4 

One obtains a better idea of what is involved by comparing Na + , Ne 
and F~. These have the same final electron configuration, i.e. are isoelec- 
tronic but the nuclear charge is 11, 10, 9 respectively. Consequently one 
would expect the diameter of Na + to be smaller than Ne and that of F~ 
to be larger. This is indeed the case, the values being 2-0, 2-5 and 2 ; 7 A. for 
Na + , Ne and F" respectively. 

2- 1 -8 Density of solid elements 

An increase in atomic number usually implies an increase in atomic weight. 
There is also some increase or decrease in atomic diameter (see Tables 3 
and 4) but it is not as large a change as the change in atomic weight. Con- 
sequently although there is generally an increase in density of the solid 
element as the atomic number increases it is not necessarily a monotonic 

It is interesting to note that if we compare the atomic weight of hydrogen 
with that of uranium (1 : 238) it is not very different from the ratio of their 
densities (007 : 18 4 g. cm." 3 ). 

12 Gases, liquids and solids 

2*2 The forces between atoms and moJecok* 

2-2- 1 Forces due to the ionic bond 

The simplest forces between atoms are those which arise as a result of 
electron transfer. A simple example is that of, say, sodium fluoride. The 
sodium atom has a nuclear charge of + 11, with 2 electrons in the K shell, 8 
in the L shell and 1 in the M shell. The fluorine atom has a nuclear charge 
of 9 with 2 electrons in the K shell, and 7 in the L shell. The outermost 
electron in the sodium atom may transfer readily to the fluorine atom; both 
atoms then have a complete shell but the sodium now has a net charge of + 1 
and the fluorine a net charge of - 1. These ions therefore attract one another 
by direct coulombic interaction. The force between them is strong - it varies 
as x~ 2 , where x is the distance between the ions, and it acts in the direction 
of the line joining the ions. Furthermore it is unsaturated - one positive ion 
can attract several negative ions around it and the force exerted by the 
positive ion on each negative ion is not affected by the presence of other 
negative ions. Of course the negative ions will also repel one another. 

2*2-2 Forces due to the covalent bond 

These are sometimes referred to as valency forces, for they are the type of force 
which accounts for the binding together of atoms in those molecules where 
the primary attraction is not ionic. For example, the force binding two 
hydrogen atoms together to form a hydrogen molecule is of this nature. 
Covalent bonding always involves the sharing of electrons. One way of 
describing this is to say that in the hydrogen molecule the single electron 
of each atom interacts with both nuclei, so that each atom regards itself as 
'possessing' two electrons which provide a complete, stable 'molecular' K 
shell. However the binding cannot be described in classical terms. It arises 
from the fact that the electrons e t and e 2 are indistinguishable, so that if 
Ci and c 2 are interchanged this does not involve a new configuration, but 
leads to an 'exchange energy' which provides the force binding the atoms 


'&M IQi 


Figure 3. Electron concentration in a pair of hydrogen atoms; (a) unstable, the 
molecule dissociates into two atoms, (b) stable, the electron concentration between 
the nuclei serves to bond the nuclei together as the H 2 molecule. 

13 Atoms, molecules and the forces between them 

together. Descriptively we may say that there is an appreciable concentration 
of electrons (negative charge) between the two nuclei (positive charge), 
and this binds the atoms together. This is indicated schematically in 
figure 3. 

Covalent forces are strong, they fall off rapidly with separation; they act 
in specified directions (valency bonds) and they are saturated. 

2-2-3 Van der Waah or dispersion forces 

Van der Waals forces exist between all atoms and molecules, whatever 
other forces may also be involved. We may describe the origin of these 
forces by considering, for example, the interaction between two noble gas 
atoms, say two helium atoms. The atoms have complete K shells; they show 
no covalent bonding or ionic bonding; also since the electron shells are 
symmetrical about the nucleus the atom has no dipole. The latter statement is, 
however, only true as a time average. At any instant there may be some 
asymmetry in the distribution of the electrons around the nucleus. The helium 
atom at this instant, therefore, behaves as a dipole of moment ft (see 

— ^-> 

< * > 

Figure 4. A dipole ft produces an electric field at a distance x along its axis of 
amount £=—5. This field can polarize another molecule such that mutual 
attraction occurs. 

figure 4). At a distance x along the axis of the dipole the electrostatic field 
E is proportional to 

~ 3 or *=-. oj, 

If there is another atom at this point it will be polarized by the field and will 
acquire a dipole n\ 

M' = ccE, (2.2) 

where a is the atomic polarizability of the atom. 
Now a dipole n' in a field E has a potential energy V given by 

V = - ft'E 

, -«k 2 n 2 

14 Gases, liquids and solids 

This implies a force between the atoms of magnitude 

- 8V constant ,, A y. 

F = = - — . (2.4) 

8x x 1 

This derivation is rather crude since, of course, the second atom will also 
react on the first. The main point of this simplified treatment is to show that 
whatever the direction of the instantaneous electron fluctuation around 
one atom, it will always produce an attractive force on a neighbouring atom. 
The magnitude of the attractive force is also dependent on the frequency of 
these fluctuations. Indeed F. London (1930) showed that the potential energy 
between two identical atoms is given by 

3 a 1 

P.E. = -~ — hv, (2.5) 

4 x 6 

where a is the polarizability, h is Planck's constant, and v is the effective 
frequency of these fluctuations. As we shall see in Chapter 9 this frequency 
is the same as that which accounts for the dependence of refractive index of 
an assembly of atoms on the frequency of the incident light, i.e. it is the 
frequency involved in optical dispersion. For this reason the van der Waals 
forces are sometimes referred to in the older literature as dispersion forces. 

Van der Waals forces are much weaker than ionic and covalent forces. 
The forces are central and like ionic forces they are unsaturated; also, to a 
first approximation, they are additive (see below). 

2-2-4 Retardation effects in relation to van der Wools forces 

The derivation of the London relation is valid only if the atoms are less 
than a few hundred angstroms apart; with larger distances the electro- 
magnetic field from one atom takes an appreciable time to reach the next 
atom. For example, if the separation is 3000 A. (3 x 10" 5 cm.), the time taken 

3x 10~ 5 

i s z tttt = 10~ 15 sec. This is comparable with the period associated with 

3xl0 10 

the electron fluctuations; consequently there is a phase lag in the inter- 
action. The theoretical treatment is very difficult but leads to the result that 
the force falls off as 1/jc 8 instead of 1/jc 7 . This 'retardation* effect becomes im- 
portant in studying the attraction between large solid bodies which are brought 
to within a small distance of one another. Since the van der Waals forces are 
roughly additive they may be integrated over the infinite half spaces. 
Allowing for retardation it turns out that for parallel surfaces, of separation 
h, the integrated force becomes proportional to \jh*. The forces themselves, 
except when h is less than say 20 A., are very small. For a separation of 
5000 A. the force is of the order of 10" 6 g-wt cm. -2 of surface. It is for this 
reason that one can safely say that the range of action of surface forces is 
small; in theory they extend to infinity, but their magnitude is appreciable 
only for a few tens of angstroms from the surface. 

15 Atoms, molecules and the forces between them 

The consequences of this may at once be seen. Surfaces will stick together 
strongly only if they can come into close contact (within a few angstroms) 
of one another. Again gases will generally be adsorbed strongly for the first 
monolayer but the attraction falls off so rapidly with distance that second and 
third layers are far less likely. Roughly speaking the formation of adsorbed 
layers depends on whether the potential energy drops by a greater amount 
than the mean thermal energy. For a molecule, the latter is of the order of 
kT, where k is the Boltzmann constant and T the absolute temperature 
(see Chapter 4). This is not the whole story since, of course, the latent heat 
of condensation is also a factor encouraging further adsorption, whilst the 
entropy decrease associated with an 'ordered' adsorbed film (see next 
chapter) tends to oppose adsorption. The net result is that polymolecular 
adsorption is unlikely unless the gas or vapour is very near saturation. 

With large particles such as those in colloidal systems or Brownian sus- 
pensions the position is different. The thermal energy of the individual 
particle is still given by kT but the van der Waals energy must be integrated 
for the millions of atoms in the particles. In such systems the van der Waals 
forces can in fact play a basic part in determining whether particles will stick 
together or not, but of course a very small amount of electrostatic charge is 
likely to swamp the van der Waals forces even for colloidal systems. 

2-2-5 Repulsion forces 

If atoms were subjected only to attractive forces all atoms would coalesce. 
Thus it is clear that for very short distances of separation some type of 
repulsive force must operate. In ionic systems the attractive force arises 
from the electrostatic charge and the repulsive force from the electron shells, 
which act as a sort of tough elastic sphere resisting further compression. 
The repulsion may be described as arising from two effects. First, the pene- 
tration of one electron shell by the other, which means that the nuclear 
charges are no longer completely screened and therefore tend to repel one 
another. The other effect arises from the Pauli exclusion principle which states 
that two electrons of the same energy cannot occupy the same element of 
space. For them to be in the same space (overlapping) the energy of one must 
be increased - this is equivalent to a force of repulsion. 

With covalent and van der Waals forces the repulsion and attraction 
are all part of a single mechanism and it is not correct to consider them as 
arising from separate mechanisms; however, for the purpose of simple 
calculation it is very convenient to do so. We therefore describe the long- 
range attraction by a term of the form 

and the repulsion by a relation of the form 
16 Gases, liquids and solids 



where B is a constant and n has a value of the order 8 to 10. In fact the 
power law relation is not very good and there are theoretical reasons for 
preferring a relation for the repulsive force of the form Be - *" where B and s 
are suitable constants. In the following calculation we shall for simplicity 
retain the power law for both attraction and repulsion forces. 
The resultant force is then 


F=— -— 





"-""It— x„— y 


Figure 5 (a) Force and (b) potential energy curves for two atoms (or molecules) 
as a function of separation. There is a long-range attractive force and a short- 
range repulsive force which operates only when the molecules come close 
together. The equilibrium separation x occurs when the net force is zero or when 
the potential energy is a minimum. 

and this is plotted in figure 5. The equilibrium separation x occurs when 
F = 0. This gives B = Ax%- m . 
Hence equation (2.8) becomes 


We may at once note that for small displacements from the equilibrium 
position the restoring force is 




17 Atoms, molecules and the forces between them 

Since the force is proportional to the displacement the motion of the dis- 
placed particle will be simple harmonic. This conclusion is of course not 
restricted to power law relations. Any relation of the form F = f(x) will 
lead to a similar conclusion. 

2-2 6 Potential energy 

The potential energy V involved in bringing one atom from infinity to a 
distance x from another atom is given by 

V = J (external force)<£* = - J (internal force)*/* = / Fax, (2.11) 

since Facts in the opposite direction to dx. 
From equation (2.9) this gives 

f-4-J L_ + _Li2z:i 

l m-\ x"- 1 n-l *"-*] 

attractive term repulsive term" 

Note that the attractive term gives a negative potential energy, the repulsive 
term a positive potential energy (see figure 6). 



iiliitiitiiiiitiiiiiiifiiiliiiiiiiiiiutiitiiiiiiiiiimiiititimmiiimmmtmiiiiiDiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiKMiiiiiiii urn ummiiiiiiiiitijii 

Figure 6. Stylized figure of potential energy curve of one molecule in relation to a 
single neighbour emphasizing the depth of the potential energy trough A£, the 
extremely steep repulsive part of the curve, and the effective diameter a of the 
molecule. The zero-point energy. ihv , where v is the fundamental frequency of 
the pair, is very small compared with AE and will be neglected in the rest of this 

At the equilibrium position where x = x the potential energy has a 
minimum value 

18 Gases, liquids and solids 

Vai *~~ A xZ-\m-\ n-\\' 
The interatomic forces are summarized in Table 6. 

Table 6 Interatomic forces 

Type Nature A m Energy to separate 

2 atoms or ions 



very large 


Na + + F"; 10" xl erg 




very large 


H-H; 7xl0" 12 erg 

Van der 


small to 


Ne:Ne; 5x10" "erg 



very small 

• The attractive force between two hydrogen atoms in the hydrogen molecule is roughly 
proportional to 1/x 3 . This only applies for small displacements; it cannot be applied over 
large displacements nor can it be considered as being valid for covalent forces in general. 

2-2-7 Intermolecular forces 

Intermolecular forces are of the same nature as those occurring between 
atoms. For ionic crystals such as rock-salt one cannot distinguish individual 
molecules of NaCl, as the crystal is an array of positive (Na + ) and negative 
(Cl~) ions held together by coulombic forces. In covalent solids such as 
diamond the individual carbon atoms are joined, by valency bonds, to form 
a strongly-linked crystal. In a solid such as paraffin wax, or solid hydrogen, 
the molecules are held together by weak van der Waals forces. However 
in a polymer these may be augmented by the attraction between polar groups 
or by chemical bonds. There is an additional type of bond which occurs only 
in the aggregate; this is the metallic bond. Individual metal atoms exist as 
such in the vapour phase, where a small number of diatomic molecules may 
occur, but there is very little tendency for the outermost valency electrons to 
join forces and produce polyatomic molecules in the vapour phase. There is 
however strong bonding in the condensed solid state. The atoms become 
ionized and exist as positive ions in a sea of free valency electrons; they are 
held together by the strong attraction between the ions and the electrons. 
In effect energy has to be provided to ionize the metal atoms, but this is more 
than compensated for by the binding energy between the ions and the 
electron sea. 

19 Atoms, molecules and the forces between them 

The simplest measure of the intermolecular forces is the heat of sub- 
limation, which is the amount of energy needed to separate all the molecules 
from all their neighbours. This is highest for ionic and covalent solids, 
high for metals and least for van der Waals solids. We may make this more 
quantitative in the following way. If the heat of sublimation E, is given in 
calories per gram-molecule, we multiply it by 4-2 x 10 7 to convert to ergs and 
divide by 6 x 10 23 to obtain the energy in ergs per single molecule. This con- 
version factor is 7 x 10" 17 . The result is the energy E, required to break bonds 
between one molecule and all its neighbours. If, therefore, the coordination 
number (number of nearest neighbours) is n and only nearest-neighbour 
interaction is involved, we may write for the binding energy E between one 
molecule and one neighbour 

E, = inE. 

The £ is introduced since otherwise every bond would be counted twice. 
We then have 

£.x7xl0- l7 = -£ or E= 14xl0 ~" E 
2 n 

As an example we quote the simplest type of solid -solid neon, which forms 
a close-packed face-centred cubic (f.c.c.) structure of the van der Waals 
type. A sketch illustrating this structure is given in Chapter 8, figure 48. The 
latent heat of sublimation is found by experiment to be approximately 450 
calories per gram-molecule. The number of nearest neighbours, n, is 12. 

E~ 5x10"" erg. 

This agrees well with the theoretical interaction energy between one neon 
atom and its neighbour. 

Another example, which is less direct, is worth quoting because it provides 
a check on our deduced value of E by a completely different method. In 
sodium fluoride the energy to convert 1 mole of NaF to Na + and F~ free 
ions is 213,000 cal. This is approximately 15x10"" erg per molecule of 
NaF or 7-5 x 10" 13 erg per ion of Na + or F~. In the crystal lattice each ion 
is surrounded by 6 ions of opposite sign (i.e. n = 6). The energy per unit bond 
is therefore 

7-5 x 10 _ " 
E = L2 ~ L ~ = 2-5 x 10"" erg. 

We may now consider the ionic interaction between a Na + and a F - as they 
are separated from their equilibrium position in the lattice (separation 
*o = 2-31 A.) to infinity. Ignoring the repulsion term, we have 

E=-\ —dx= = -10x10"" erg. 

20 Gases, liquids and solids 

This assumes only nearest-neighbour interaction. There is, in fact, a very 
large interaction between the next nearest neighbours. For example, the 
potential energy due to the attraction between one positive ion and the 
negative ions surrounding it, is greatly reduced by the coulombic repulsion 
between the positive ion and the nearest positive ions. These are only a 
little further away than the nearest negative charges and as coulombic 
potentials fall off only as l/x, the resultant potential energy is greatly reduced. 
Detailed calculations such as those indicated in Chapter 10 show that the 
energy is reduced from —e z /x to 0-2905 times this value. This gives a value 
for E of about 2-9 x 10~" erg which agrees well with the value calculated 
from the heat of sublimation. 

The main features of intermolecular forces are summarized in Table 7. 

Table 7 Intermolecular forces 




Heat to sublime solid to 


erg per bond xlO 1 * 


identical with 


break up into 




vapour of NaF 




giant molecule 
held by covalent 


break up into 
C atoms 


Van der 

as for atoms 

solid methane 

break up into 






metal ions in 
sea of free 
valency electrons 


vaporize to 
metal atoms 


2-2-8 Comparison of electrical and gravitational attraction 

Consider the interaction between, say, two helium atoms where only the 
weakest electrical attraction exists as a result of the van der Waals mech- 
anism. If the atoms are packed together as close as possible (say in liquid 
state) the separation x a is about 2 A. The potential energy V w due to van der 
Waals for this separation turns out to be about — 10" 1S erg. Now consider 
the potential energy V t due to gravity. 


J x Xo 

6-7xl0- g (0-7xlQ-") 2 
2xl0- 8 

21 Atoms, molecules and the forces between them 

i.e. V, = -10~ 46 erg 

whereas V w = -10~ 15 erg. 

We see that even when only van der Waals forces are involved the gravi- 
tational energy is trivial compared with the electrical. 

2-2-9 The simplified potential energy curve 

In this section we shall treat molecules as though they are hard elastic balls 
of diameter a which attract each other according to some suitable law of 
force, and then experience a very powerful repulsion force for separations 
less than a. 

The resulting potential energy curve is then as shown in figure 6 and the 
two most important factors are a and the depth Ae of the potential energy 

Some typical values are as follows in Table 8. 

Table 8 


a A. 

Ae erg (approx.) 



lxlO- 15 

H 2 


4xl0~ 15 



5xl0" 15 

N 2 



co 2 


40xl0- 15 

In later chapters we shall be able to explain, in terms of such a potential 
energy curve, a large number of phenomena. These include critical temperature 
of a gas, surface tension and viscosity of a liquid, heat of sublimation of a 
solid, elastic constants and the thermal expansion of a solid. 

Postscript on electronic energy levels 

The fact that the simple kinetic theory of gases is valid sheds light on the electronic 
energy levels in the atom. Consider for simplicity a monatomic gas. The whole of 
the gas behaviour is consistent with the view that the only energies involved are the 
kinetic energy (%tK) of the atom as a whole (see Chapter 5). Collisions do not share 
the kinetic energy with say the electronic energy of electrons within the atom. The 
reason (see N. F. Mott, Contempory Physics, vol. 5, 1 964, p. 401 ) is that the electronic 
energies are quantized and the gap between the various energy levels is enormous 
compared with the ordinary kinetic energy due to thermal motion (5 eV compared 
with 0-02 eV at room temperature). For the same reason the electronic energies play 
no part in the specific heat. Ordinary temperatures are far too small to promote the 
electrons from one energy level to another. This has a direct bearing on the Langevin 
treatment of diamagnetism (see pp. 258, 269). 

22 Gases, liquids and solids 

Chapter 3 

Temperature, heat and the laws 

of thermodynamics 

In this chapter we consider briefly some of the basic concepts of heat and 
temperature. The treatment is simple and restricted to those aspects which 
will be of use in later parts of the book. 

3-1 Temperature 

3 • 1 • 1 The concept of temperature 

The concept of temperature originally arose from a sensory feeling of hot 
or cold. Probably all physical concepts arise in a similar way; the next step 
is to turn the idea into something more general and, in particular, more 
objective. One such approach is as follows. It is found that a given mass of 
gas is completely specified by its volume V and its pressure P. Suppose we 
start with 1 kg. of a gas and subject it to any changes we wish whether by 
compressing it, cooling it, allowing it to expand heating it; we then take 
another 1 kg. specimen of the same gas and carry out a completely different 
series of operations. If we end up with the same values of V and P for the two 
samples, the final states will be found identical in every way - colour, warmth, 
viscosity, and every objective and subjective test to which we can subject the 

If we end up with different P, lvalues there is something different about the 
samples and if they are placed in contact changes take place until they reach 
identical P, lvalues (see section 3-1-3 below). 

The attribute which produces this difference is the temperature and the 
changes brought about when the specimens are placed in contact result in 
their eventually reaching the same temperature. For gases and fluids there is a 
unique function of (/>, V) which determines their temperature 6 if the mass is 
specified: /(/>, V) = 6. For solids one needs more parameters. 

3-1-2 Temperature scales 

There are, of course, sophisticated means of deriving an absolute thermo- 
dynamic scale of temperature (see section 3-3-6 below). In practice one can 
use the P-V properties of a gas. One may also use less direct standards such 
as the expansion of a metal rod, the expansion of a liquid column (a con- 
ventional thermometer), or the e.m.f. produced by a thermocouple. 

23 Temperature, heat and the laws of thermodynamics 

3-1 -3 Zeroth taw of thermodynamics 

In the development of thermodynamics the first two laws were developed 
and so named, before it was realized that a preliminary law was required to 
satisfy a rigorous formulation of thermodynamic laws. This was therefore 
called the zeroth law. (It has nothing to do with Professor Zero.) The law 
simply states that if two bodies A and B are individually in thermal equili- 
brium with a body C, then A and B are in thermal equilibrium with one 
If C is a thermometer this is clearly a matter of common experience. 

3-2 Heat 

3-2-1 The concept of heat 

Suppose we have a body of mass Mi at temperature lt which is placed in 
thermal contact with another body of mass M 2 and temperature 2 , where 
2 is greater than 1 . Equilibrium will be reached at a common temperature 
6,6i> > 0l We say that heat has flowed from Z to U 

Without specifying what heat is, we can define its units. The heat required 
to raise the temperature of 1 gram of water at 15°C. by 1°C. is the calorie. 
If we assume that in any experiment involving the flow of heat from one 
body to another heat is conserved we may define the specific heat of a given 
substance in terms of the specific heat of, say, water. The specific heat s is 
defined as the number of calories required to raise the temperature of one 
gram of the substance by 1°C. 

We can now carry out experiments with different masses of different 
materials at various temperatures using only a centigrade scale thermometer. 
Common experience then shows that all the results are consistent with our 
original assumption that heat is always exactly conserved. We have to 
qualify this by adding that this is true only if any volume changes which occur 
do a negligible amount of work. 

So far we have said nothing about the nature of heat, but this does not 
affect the validity of the calorimetric conservation principle which we have 
just described. The next step is to consider briefly J. P. Joule's classical experi- 
ments which show that heat is, in fact, a form of energy. 

3-2-2 Joule's experiments 

J. P. Joule (1818-1889) began his famous series of experiments on the relation 
between heat and energy in 1840 and summarized his major conclusions in a 
massive m6moire published in the Philosophical Transactions in 1850. His 
first experiments (1843) involved the combined effects of mechanical and 
electrical energy, while his second experiments (1 845) were concerned with the 
heat liberated when a gas is compressed. His more famous experiment in 
which he directly compared the mechanical work expended in stirring 

24 Gases, liquids and solids 

water with the heat evolved did not appear until later in 1845, and a greatly 
improved version was described by him in 1878. The principle was to stir 
water in a copper vessel with paddles and to calculate the mechanical work 
done W. Using a very sensitive thermometer he observed the temperature 
rise At of the water, paddles and vessel, and estimated heat losses by con- 
duction, radiation and (if relevant in the mechanical arrangements of the 
driving system) the losses due to friction in the bearings. From subsidiary 
calorimetric experiments he was able to determine the specific heats of the 
vessel and the paddles in terms of the unit heat capacity of water. In this way 
he could calculate the total amount of heat H in a purely calorimetric 
experiment required to produce the same temperature rise as in the paddle- 
stirring experiment, viz: 

H = At [Mass of water + rtiiSi (vessel) + m 2 st (paddles)] 

+ losses. (3.1) 

He then found that H was closely proportional to the mechanical work W. 
We may summarize some of his other experiments. 

(a) He repeated the stirring experiments with mercury as the liquid. 

(b) He rubbed iron rings together against their interfacial friction. 

(c) He passed an electric current through a resistance wire immersed in a 

(d) He compressed air in an insulated cylinder. 

In all cases he found that the observed temperature rises could be achieved 
simply by adding heat of an amount proportional to the mechanical or 
electrical work done. Further, the factor of proportionality was a constant 

W=JH (3.2) 

where / is Joule's constant or the mechanical equivalent of heat and has 
the value/ = 4-187x 10 7 erg calorie -1 . 

Joule's constant is of immense importance in all mechanical-thermal 
operations, but in effect this reduces to translating specific heat data into 
energy data. It is clearly not a fundamental unit of physics. Indeed if calori- 
metric studies (involving pounds or grams of water and degrees centigrade 
or fahrenheit) had not been so attractive as a convenient means of describing 
thermal energy, it is possible that our basic unit would have been the Joule 
(10 7 erg); we should then have had a calorimetric heat-unit defined as the 
heat required to raise the temperature of one gram of water by 0-238°C. 

3>3 The laws of thermodynamics 

3*3*1 Internal energy and the first law of thermodynamics 

If we carry out an experiment in a thermally isolated calorimeter, such that 
no heat can flow into or out of the system, we can change the state of the 

26 Temperature, heat and the laws of thermodynamics 

system by performing mechanical or electrical work on it. Joule's experiments 
show that for the same amount of work we shall always arrive at the same 
final state. We may now define the difference in internal energy U of the 
system as the work done on the system in an adiabatic calorimeter. 

Since we can change from state 1 to state 2 by an infinite number of 
combinations of heat and work, if we wish to define internal energy as a 
function of state only, we must include both work and heai in our definition 
of energy. The following example illustraies this in a more concrete way. 

Suppose we take a mass of water and perform mechanical and/or electrical 
work on it ; provided we do the same amount of work W, we produce the same 
change in temperature. Alternatively we can apply a quantity of heat H- W\J 
and produce the same change. If we start with water ai 0°C. and end with 
water at I00°C, then the water ends with the same energy state whatever the 
path adopted. For example, we could add heat to raise the temperature to 
30°C. and then stir paddles to supply the remaining 70°C. rise. Clearly an 
infinite number of paths are available. Yet the change in internal energy 
C^ioo - ^o (ignoring any work done by the liquid on expanding) must be 
determined only by the initial and final temperature, not by the path. If this 
were not true we should be able to devise heating and cooling cycles whereby 
we could raise the temperature to 100°C by one path and cool it back to 
0°C. by another, but still obtain energy from the system. This is contrary to 
Joule's experiments. 

We are left with the conclusion that a change in U <s a function only of the 
initial and final states. If we do external work AW and add heat AQ we have 

AU=AW+AQ, (3.3) 

where A 17 is uniquely defined, but Aff and AQ can have any values provided 
their sum is equal to AU. 
For a specified change in state 

W A + Q A over path A = W B + Q e over path B. (3.4) 

Thus work plus heat are conserved. We may regard this as an extension 
of the conservation of energy which is well established in mechanics and 
conclude that: 

Energy is conserved if we include heat as one of the forms of energy. 

This is one way of formulating the first law of thermodynamics. 

In the simple example given above we have considered U to consist of heat 
and mechanical or electrical energy. A little consideration shows that the 
internal energy can embrace all forms of energy, thermal, mechanical, 
electrical, gravitational, electromagnetic, etc. It may again be shown by 
similar arguments that, however many types of energy are involved, any 
change in the internal energy does not depend on the path; it is a function 
solely of the initial and final states. 

26 Gases, liquids and solids 



Internal energy of fluids 

For a fluid, U is found to be a unique function of only 2 variables; either 
(/», V), (/>, 7") or (V, T), where P is the pressure, V the volume and T the 
temperature. Since U is independent of path, changes in it can always be 
expressed in terms of an exact differential. This also means that, in principle, 
we can integrate dll between initial and final states and always end up with 
the same answer. Thus we can always write 

for U(T, V) 




for U(P. V) 

\dPjr \dV) P 

In our example of raising the temperature of water from 0°C. to 100°C. we 

concentrated on the variation of U with temperature; ( — ; J in equation 

\BT/ V 

(3.5a). That is why we made the proviso that for this to be strictly true any 

changes in volume would have to be ignored, i.e., ( — j = 0. 


Reversible changes in a gas 

If we allow a gas to expand, it does work against its surroundings. The work 
done depends on the pressure of the gas, the volume change and the pressure 
outside the gas against which it does work. 

Consider a cylinder of cross-sectional area A closed with a weightless 
piston. The work done during an expansion can be denned uniquely only if 

Figure 7. Reversible expansion of a gas. 
27 Temperature, heat and the laws of thermodynamics 

(a) the piston is frictionless, (A) the expansion is slow and always in quasi- 
equilibrium, i.e. the pressure outside the piston is only infinitesimally smaller 
than P so that, by a minute increase in the external pressure, we would be 
able to compress the gas. Under these conditions the gas pushes with a force 
PA against the surroundings and if the piston moves a distance dx the work 
done is PAdx = PdV, where d Vis the volume increase. For such & frictionless, 
reversible expansion PdV is uniquely defined for the increment dV. 

Consider now what happens if we increase the internal energy of a gas by 
allowing it to do work w, and also add heat of amount q. We have 

unique not unique not unique 

dU = -w + q. (3.6) 

Here w is the work done by the gas and is therefore counted negative. For 
reasons, which we described before, dU is determined uniquely by the 
initial and final states. This is not generally true of — w and q separately and 
for this reason they cannot be expressed as exact differentials. 

If we now specify that the work w is a reversible frictionless expansion 
we have 

unique unique 

dU = -PdV +q. (3.7) 

Consequently q must also be uniquely defined over this increment of change 
so that it can be written as dQ. 

dQ = dU+ PdV. (3.8) 

Each of the quantities in equation (3.8) is now in the form of a differential, 
so if the relevant paths are known, they can all be integrated exactly. We see 
now why frictionless reversible expansions or compressions are so important 
in thermodynamics. In an electrochemical system the term equivalent to 
PdV would be Edq where E is the potential difference and dq the amount of 
charge passing. Here again for an equation like (3.8) to be valid the current 
flow would have to take place against an opposing potential infinitesimally 
less than E. In a magnetization process, where a magnetic element of moment 
Mis exposed to a magnetic field dH, the equivalent term would be MdH. 

3-3-4 Enthalpy 

We see from equation (3.8) that the internal energy may be written 

dV=dQ- PdV. (3.8a) 

Unfortunately we cannot integrate this, unless we know the equation of state 
and the path followed by the expansion cycle. For example an adiabatic 
expansion would give a value of J PdV which differs from that occurring 

28 Gases, liquids and solids 

in an isothermal expansion. If, however, we specify that the expansion is 
maintained at constant pressure we have 

U 2 -U t = Q 2 -Qx-P{V 2 - Fi). (3.9) 

The heat absorbed at constant pressure is then given by 

Qi-Qi = (U 2 +PV 2 )-(U 1 +PV l ). (3.10) 

We now introduce a new thermodynamic quantity called enthalpy which is 
defined as 

H = U+PV. (3.11) 

This is also a function solely of initial and final states. Thus for a small 

dH= dU + PdV + VdP = dQ+ VdP. (3.12) 



As we shall see later enthalpy is conserved in the steady flow of a gas through 
a throttle (Joule-Kelvin expansion). 

3-3-5 Specific heats 

These may be defined as the amount of heat required to raise the tempera- 
ture of a specified amount of material (one gram, one gram-molecule) by a 
unit degree of temperature: 

C = ^. (3.14) 


In general we can specify two different conditions for the measurement of C. 
Constant volume 

C " = (fl = (I) v fr0mCqUati0n(3 ' 8a) - 

= (fl=(f). 

For gases, as we shall see, there is an appreciable difference between C P 
and C v . For liquids and solids the difference is very much smaller. Some 
typical values of C for metals are given in Table 9. 

29 Temperature, heat and the laws of thermodynamics 

Constant pressure 
C P 

Table 9 Specific heat of some metals 



heat C 



cat. g.~ l 

"Cr 1 


















The last column shows that the product of the specific heat and the atomic 
weight (this is in effect the specific heat per gram atom) has a value of about 
6cal. g-atom -1 °C. -1 . This conclusion is known as the law of Dulong and 
Petit and we shall discuss its basis and validity in a later chapter. 

3 -3 -6 The second law of thermodynamics 

It is not the purpose of this book to provide a complete account of thermo- 
dynamics (readers will find a fascinating and stimulating account in A. B 
Pippard (1957), Elements of Classical Thermodynamics, Cambridge University 
Press). Indeed we shall be able to deal with practically all our needs using only 
the first law. There is, however, some point in emphasizing the essential 
difference between heat and other forms of energy. When a force does work, 
the work done is the product of force and distance moved, the force always 
being measured in the direction of the line along which it is acting. When an 
electric current does work the current flows in the same direction as the 
resultant e.m.f. Mechanical work and electrical work are in principle absolu- 
tely convertible into one another. A dynamo in principle produces an amount 
of electrical power exactly equal to the work done in driving it (apart from 
friction and electromagnetic losses). If this is converted into heat there is, 
again, an exact equivalence as Joule showed. If, however, we wish to convert 
heat into work this is not true, because thermal energy always has associated 
with it random movement, vibration, thermal agitation, so there is always 
part of the thermal energy which is not free for conversion into mechanical 
or electrical energy. 

N. L. S. Carnot (1824) showed that if one had a heat engine working between 
a source at a higher temperature and a sink at a lower temperature, the engine 
would achieve its greatest efficiency if all the parts of the cycle were carried 
out reversibly. Under these conditions the efficiency was independent of the 
working material. By efficiency we mean the ratio of the useful work done 
to the thermal energy absorbed from the higher temperature source. For 

30 Gases, liquids and solids 

reversible cycles if the heat taken in is Q t and the heat rejected to the sink is 
Q 2 , the efficiency r\ is 


Since this is independent of the working material it depends only on the 
temperature of the two sources. This provides a means of establishing a 
thermodynamic scale of temperature, which is achieved by denning the ratio 
of heat absorbed at the higher temperature to that rejected at the lower 
temperature as equal to the ratio of the higher to the lower temperature, 

O T 
i.e., — = rr- For such an absolute scale of temperature 

Qi Tz 

}£ -^T 1 = ~- 1 - (3.17) 

We see that fundamentally the efficiency can never be unity unless T 2 , the 
temperature of the sink, is zero. Under these conditions the heat, as it were, 
is acting completely in the direction of doing useful work. 
For the reversible cycles we see that 

£-4". (3-18) 

Hence travelling from the initial condition, through the cycle, and then back 
to the initial condition again 

Qt Q* 

|r-|r=0. (3.19) 

We may now define dQ/T as the entropy change dS in a reversible process. 
The word entropy was invented by R. Clausius ; it comes from the Greek word 
trope meaning transformation and he deliberately added the prefix en to 
convey a connection with energy. For a reversible cycle (the sign § indicates 
that the integral is taken over the complete cycle) 



This implies that dS is an exact differential; for any increment in a reversible 

dS^^Q or dQ = TdS. (3.21) 

One of the most interesting properties of entropy is the following. Even 
for a reversible cycle the integral of dQ depends on the details of the paths 
followed (see discussion under enthalpy). If, however, we divide the increment 

31 Temperature, heat and the law of thermodynamics 

of dQ by the temperature at which the heat dQ is absorbed (or rejected), 
the quantity dQ/T in any reversible path can be integrated between the initial 
and final state and is independent of the path followed. This is what is meant 
by saying that dS is an exact differential. 

For completeness we quote two other thermodynamic functions which are 
independent of path, the Helmholtz free energy A for changes at constant 
volume, and the Gibbs free energy G for changes at constant pressure. They 
are defined as follows: 

A = U-TS (3.22) 

G = H-TS. (3.23) 

We see that for an isothermal change 

dA = dU-TdS. (3.24) 

The change dA in free energy A is equal to the change in internal energy U 
less the 'unavailable' energy TdS. This is the maximum work that the system 
can perform. For a non-reversible process the available work will be less than 
dA. At constant temperature and constant pressure the maximum work 
that the system can perform is dG; this corresponds to the change in enthalpy 
less the unavailable energy TdS. 

It is not surprising, since the unavailability of thermal energy is associated 
with its randomness, that entropy should in some way be connected with the 
probability or distribution-randomness of the system under consideration. 
What is surprising is that this concept can be made quantitative. If the 
probability of rinding a system in a given state is W t and in another state is 
W 2 , the entropy change may be written 

AS = kln^, (3.25) 

where k is the Boltzmann constant. Of course the probability needs to be 
correctly defined and this forms the subject of statistical mechanics. The 
probability associated with the translational motion of gas molecules turns 
out to be proportional to the volume occupied by the gas; the larger the 
volume the greater the randomness. The entropy change in increasing the 
volume from Vi to V 2 at constant temperature is 



where N is the number of molecules in the quantity of gas considered. We 
shall use this result in a later chapter of the book. 

32 Gases, liquids and solids 

Chapter 4 

Perfect gases-bulk properties 

and simple theory 

4-1 Bulk properties 

4*1*1 Summary of main bulk properties 

As we shall see in the course of this and the next two chapters, all gases will 
approximate to perfect or ideal gases if they are sufficiently dilute and if the 
intermolecular forces are negligible compared with the thermal energy. This 
condition applies reasonably well to oxygen, nitrogen and hydrogen at normal 
temperatures and pressures. For this reason the earliest studies of the 
'springiness' of air were concerned with the behaviour of a perfect gas. 

The main bulk properties were established long ago. We summarize the 
main conclusions: 

1. Boyle's law (co. 1660). For a given mass of gas at a fixed temperature 
the product of pressure and volume is a constant. 

PV = constant. (4.1) 

2. Charles' law (1787) or Gay-Lussac's law (1802). For a given mass of gas, 
if the pressure be kept constant the volume increases linearly with the 

V= K (l + a/). (4-2) 

If the temperature is measured on the centigrade scale and V is the volume 
at 0°C, it is found that a has the value « = ^73- We may therefore write 
equation (4.2) as 

\ 273/ 273 

where Tis a scale of temperature, which has — 273°C as its zero point. This 
scale turns out to be the same as the thermodynamic scale. 

3. Variation of pressure with temperature if the volume is kept constant. 
P = Po(.l+fit). (4.4) 

33 Perfect gases - bulk properties and simple theory 

The constant has the same value of J73 so * at we nave 

P=^-T. (4.5) 


4. General gas equation. Rewriting equation (4.1) in the light of equations 
(4.3) and (4.5) we obtain a general equation of state 

PV = RT, (4.6) 

where R depends only on the quantity of gas used. 

5. Avogadro's law. For all perfect gases R has the same value if the same 
molecular quantity of gas is considered. For one gram-molecule 

R ~ 8xl0 7 erg°K.- 1 

as 2cal.°K.-- 

6. Dalton's law. This states that, at a fixed temperature, the pressure of a 
mixture of gases is equal to the sum of the pressures which would be exerted 
by each gas separately, if the other constituents were not there. 

4"T2 Reversible isothermal expansion 

We allow a gas to expand in a frictionless cylinder against a pressure 
infinitesimally less than the gas pressure. The temperature T is maintained 
constant. The gas does work 

dW = PdV. (4.7) 

For an ideal gas PV = RT so that P in equation (4.7) can be replaced by 

V ' 

Then dW=—dV. 

If expansion occurs, from volume Vi and pressure P x to volume V 2 and 
pressure P 2 , the work done is 

W = RTla — = RTln — . (4.8) 

V, P 2 

This work is provided by the heat absorbed from the constant temperature 
source, which must be included in the system if T is to be kept constant. 

4*1.3 Fast adiabatic expansion into a vacuum 

Consider the experimental arrangement shown in figure 8. A volume V of 
gas is maintained in the left-hand flask at pressure P and temperature T. 

34 Gases, liquids and solids 

Figure 8. Fast adiabatic expansion of a gas into a vacuum. Jojjle found that for 
dry air there was no net temperature change. 

The right-hand flask is evacuated. The whole system is thermally insulated 
so that heat cannot flow into or out of the flasks, then the tap is opened and 
the gas rushes into the empty flask. Some inequalities of temperature occur 
initially but a steady state is quickly reached. Joule found that for dry air the 
temperature in the steady state is indistinguishable from the original 
temperature. We write 

dU = &.Q+WV. (4.9) 

For an adiabatic process there is no heat flow, AQ = 0, and for an expansion 
against zero pressure no work is done, i.e. Aff = 0. 

dU = 0. (4.10) 

This means that U is unchanged and is independent of volume occupied 
(or of pressure exerted). This implies that there are negligible forces between 
the gas molecules; otherwise the internal energy would depend on how far 
we had separated the molecules, i.e. on the volume. 
Analytically we may write 

U = f(V,T) or (P,T) 

\WJ T \bT)y 

Since there is no temperature rise, dT = 0; also from equation (4.10), 
dV = 0, so consequently 

(if) r = °' andSimilarly (l) r =:0 - (4 - 11} 

Thus for a perfect gas U is a function of Tonly and neither P nor Kis involved. 

4 • 1 -4 Specific heat of a gram-molecule of a perfect gas 

Consider a small equilibrium change in the gas if we increase its temperature 
by dT. The basic equation is 

35 Perfect gases - bulk properties and simple theory 

dQ = dU + PdV 

t gas thefi 

For a perfect gas the first term on the R.H.S. is zero. Thus we have 

dQ= \^\ dT+PdV. (4.12) 

Consider now two different conditions. 

(a) Constant volume: dV - 0. We can rewrite equation (4.12) as 



(6) Constant pressure: The gas expands during heating. Equation (4.12) now 

dQ = CydT + PdV. 


For a perfect gas PV = RT so that PdV + VdP = RdT. At constant 
pressure dP = 0, consequently 

So equation (4.14) becomes 

C K + R, 
or C, - C v = R. (4.15) 


4-1 -5 Reversible adiabatic expansion 

We start with our basic equation 

dQ = dU + PdK, 

specifying that, for adiabatic process, dQ = 0. Our starting equation then 

= CydT + PdV. (4.16) 

For a perfect gas, the gas equation always applies: 

PdV + VdP = RdT. 

36 Gases, liquids and solids 

Equation (4.16) becomes : 

- CydT = PdV = RdT - VdP 
= (C P - C v )dT- VdP. 
Hence CpdT= VdP. 
From (4.16) 

C v dT = - PdV. 
Taking the ratio of equations (4.17) and (4.18) 



«fF dP 



Integrating from initial to final conditions 
K P 

- b £H 

Since the antilogarithm of is 1 this gives 

PiVl = P 2 iq = constant. 

If we plot the pressure against the volume we see (figure 9) that the adiabatic 
curve is steeper than the isothermal. 



Figure 9. Pressure-volume relations for a perfect gas. The adiabatic (lower) 
curve is steeper than the isothermal (upper) curve. 

37 Perfect gases - bulk properties and simple theory 

4-1 -6 Work done in reversible adiabatic expansion 

The work done by the gas in expanding from Pi V t to P 2 V 2 is 

dW = PdV. (4.21) 

At any stage we can write 

PV = P t VI 

so that P in equation (4.21) can be replaced by PiYlV"'. The equation 

W= | P.VJV-'dV. 

= f * P t V\V-'a 

4- 1-7 Cooling as a result of adiabatic expansion 

The work done by the gas is done at the expense of the internal energy. 
Instead of calculating this de novo we merely combine the adiabatic expansion 
equation with the general gas equation: 

PiV\ = P 2 Vl, 

^L=^. (4.23) 

T t T 2 

Taking the ratio of the L.H.S. of the two equations and equating it to the 
ratio of the R.H.S. we obtain 

P X V\ 

p 2 vi 


P 2 ^2 


T 2 

Hence p = (£\ ~ = fc\ ' • (4.24) 

4* 1-8 Entropy changes on expansion 

When a gas expands isothermally the molecules are provided with a greater 
volume without any other change. This implies a greater randomness or 
probability and this, in turn, implies an increase in entropy. 

For an isothermal expansion the entropy change for a gram-molecule 
of gas is easy to calculate, though the proof given here cannot be considered 
very rigorous. If the initial volume is V x and the final volume V 2 and there 

38 Gases, liquids and solids 

are N molecules, the probability of finding one molecule in the final volume is 
— times greater than that of finding it in V y . For the N molecules the 
relative probabilities of finding all the molecules in V 2 compared with V, is 
I — I • The change in entropy is then 

probability 2 

AS = kin' 



- w K:) -*■■(£)• «•*> 

if we are dealing with a mole of gas, i.e. if N = Avogadro's number, No. 

For the adiabatic expansion we may divide the process into two independent 
reversible parts. We first allow the gas to expand isothermal !y from K, to V 2 , 
giving the entropy increase derived above. We then cool the gas (at constant 
volume V 2 ) until its temperature is reduced from 7", to the final (adiabatic) 
temperature T 2 . The decrease in entropy for such a reversible cooling is 


But as we saw in equation (4.24) ~ for the adiabatic expansion is equal to 

I — I . The entropy loss on cooling is therefore 

C„lnJ = CV(y- 


= (C P -CV)ln 

= Rlnj±= -Rlnj*. (4.25b) 

If we add this to the entropy change associated with the isothermal expansion 
part of the process [equation (4.25)], we see that the total entropy change is 
zero. This is because the volume increase tends to increase the randomness 
whereas the temperature decrease tends to reduce it, the two effects exactly 
balancing out. This is, of course, what we should expect, since in the adiabatic 
process itself no heat is added or withdrawn from the gas, so Ag = and 

therefore AS = I — must be zero. 
J T 

39 Perfect gases - bulk properties and simple theory 

4-1 *9 Adiabatic lapse rate 

The pressure of the atmosphere diminishes with height and in the next 
chapter we shall explain why. The atmospheric pressure drops by about ^th 
part for every 1000 ft of ascent. If at ground level the pressure is P , the 
pressure 1000 ft up is about f§ P . 

Consider now what happens if some disturbance initiates an upward 
movement of air. The air will expand and in the course of this it will cool. 
Assuming the cooling to be purely adiabatic, ground temperature to be T , 
and the temperature at the 1000-ft level to be T, we have from equation 





since y for air is approximately 1 -4. If T = 300°C. 

T= 300 1 1 — — ) = 300-3, 
I 210/ 

so that the temperature drop is 3°C. This is close to the observed temperature 
drop per 1000 ft. In practice if there is an appreciable amount of moisture 
present in the air, cooling leads to condensation and this releases energy 
which lessens the cooling. A more representative value is about 2°C. per 
1000 ft. 



-60 —40—20 
temperature °C. 

inversion point 

temperature °C. 


Figure 1 0. The decrease of temperature with height in air. (a) steady conditions, 
(b) early morning. 

Figure 10(a) shows the variation of temperature with height. During the day 
the temperature drops from about 20°C. at ground-level to -60°C. at a 

40 Gases, liquids and solids 

height of 7 or 8 miles, where the troposphere ends. Often in the early morning 
the temperature gradient is reversed near the ground level due to the cooling 
of the earth overnight [figure 10(6)]. Inversion temperatures of this type can 
be an important cause of fog and smog. 

4-2 Elementary kinetic theory of ideal gases 

The idea that gas molecules are in random motion was considered by the 
Greeks [as Lucretius' description of Brownian motion shows (see the end 
of Chapter 5)]. It was revived in a surprisingly modern form by D. Bernoulli 
(1700-1782) in 1738, but seems not to have been widely adopted until J. C. 
Maxwell (1831-1879) took the matter in hand and provided practically the 
whole of our present theory of gases. We first present a very simple theory. 
Its basic assumptions are: 

(i) the gas consists of identical molecules of mass m ; 
(ii) they have zero size, and do not collide with one another; 
(iii) they exert no forces on one another; 

(iv) they undergo random motion within the gas and their collisions with 
the walls of the container are perfectly elastic. 

We now consider the collisions of the gas molecules with the container 
walls and show that the pressure exerted by the gas on the walls is due to the 
momentum transfer accompanying the collisions. 

Suppose there are n molecules per c.c. They all have varying velocities 
but because the numbers are generally so enormous we can subdivide these up 
into groups of n u n 2 , n 3 • . • molecules per c.c. with velocity ranges c x to 
Ct + dc; c 2 to c 2 + dc; c 3 to c 3 + dc, etc. 

Let the gas be held in a cubic container of side /, and let us first consider 
the group with the velocity Ci - there are nj 3 of these in the container. 

We resolve ci into three mutually orthogonal components iii, p lf w u 
parallel to the sides of the cube. Clearly 

u\ +v\ + w\ = c\. (4.2fi) 

Along the u x direction the molecule has momentum rmit normal to the face 
of the cube before collision and momentum -mui after collision. Thus for 
each collision 

momentum transfer to wall = 2mui. (4.27) 

The molecule has to travel across the cube and back again (i.e. a distance 

of 21) to make the next collision. The time taken is — . 


Hence the molecule makes — collisions per second. (4.28) 


41 Perfect gases - bulk properties and simple theory 

The momentum transfer per molecule per second is the product of equations 
(4.27) and (4.28), i.e. 

2nui, h) = ^ . (4.29) 

\2l) I 

Consequently for all the molecules of this group the total momentum 
transfer per second is 

m , 


We now define the average value of u \ for all this group as 

so that equation (4.30) becomes 

-n l Pl? 1 = mnyj 2 . (4.32) 

The momentum transfer per second is the force and this is exerted on a face 
of area I 2 . These molecules therefore exert a pressure p on the wall of amount 

p = mn^i. (4.33) 

We now make the following observation. For this group of molecules the 
velocity C\ is the same but their paths are uniformly distributed in all 
directions; when averaged over the large number of molecules involved 
there can be no preferred direction so that 

If = 7* = "^ = icj. (4.34) 

Equation (4.33) becomes 

p = ^wi.cj. (4.35) 

For all the groups of molecules the total pressure becomes 

p = 1 \mn i cf 
— f I \mrij cf = | (total kinetic energy per c.c). (436) 

For the total volume V we have 

Py = f (total kinetic energy of molecules in the volume V). (4.37) 

We now define the mean square velocity of all the molecules in the gas as 

•p _ «■<•? + "jC 2 2 + n 3 c I ... 
«i + it + • • • 

= -^ . (4.30 

42 Gases, liquids and solids 

We can insert this in equation (4.36) and obtain 

P = \ Hmn t cf = \rruK 1 . (4.39) 

For a gas occupying volume V, where the number of molecules is N where 
N = n V we have 

PV = l/nAfp. (4.40) 

For a gram-molecule of a gas N is Avogadro's number, N , and 

i»F=imAf P = iJr (4.41) 

Order of magnitude o/(c 2 )*. The square root of ~? is called the root mean 
square velocity. As we shall see below it is a little less than the arithmetic 
mean of the velocities but not very different. From equation (4.41) 
RT = ^mN u c u = j/Wt^ where M is the molecular weight. Consequently 

c> = ~. (4.42) 

Typical values for (?)* are given in Table 10 for gases at 0°C. (T = 273°K.). 
Table 10 Typical values ofr.m.s. velocity 


r.m.s. velocity 

m. sec.' 1 

H 2 




o 2 




Benzene vapour 


Mercury vapour 


Electron gas 


4-3 The ether theory of the ideal gas 

Before congratulating ourselves on the validity of the kinetic theory we may 
consider one of the theories which preceded it, and I am much indebted to 
Professor Eric Mendoza for pointing it out to me. This theory was based on 
the Newtonian view that a gas was like a solid. In a solid the molecules are 
held together by attractive forces. In a gas they are repelled by some agency 
which pushes them apart but still leaves them in a, more or less, uniform 
array. What pushes them apart ? One view was that it was the swirling ether. 

43 Perfect gases - bulk properties and simple theory 

If each molecule is surrounded by a shell of ether of radius r rotating at a 
very high speed, the centrifugal force exerts a pressure on the neighbouring 
sphere and so keeps the molecules apart. It is as though the molecules are 
surrounded by balloons which press against one another. If the ether has a 
mass m and the velocity is somehow in all directions, but confined to the 

great circles of the sphere, the net centrifugal force is — and it is exerted 


over an area 4nr 2 . The pressure exerted on the balloon shell is therefore 

_ r _ mv 2 _ h 

4itr 2 Anr 3 Anr 3 

The numerator on the right-hand side is § times the kinetic energy of the ether ; 
the denominator is the volume occupied by the ethereal shell. We are left 
with the relation 

p = § (kinetic energy of ether per unit volume). 

This is very similar to equation (4.36) and shows that the right answer does 
not necessarily guarantee the validity of the physical model, a most salutory 
warning. There is, of course, more to this than coincidence. Any gas theory 
which attributes the gas pressure to the kinetic energy in the gas is bound to 
give a relation similar to that obtained in the kinetic theory. This follows 

because pressure has the dimensions of and energy per unit volume has 


.. .. -(force x distance) .... , force 

the dimensions of which is also . 

volume area 

4-4 Some deductions from kinetic theory 

4-4-1 Thermal equilibrium between gas and container 

We now make use of a treatment described by J. Jeans in his Kinetic Theory 
of Gases to analyse the conditions which determine the thermal equilibrium 
which must exist between the molecules of a gas and the molecules of the 
container. We start off by considering the collision between one gas molecule 
and one wall molecule. Let m, M be the mass of the gas molecule and wall 
molecule respectively, and let their velocity components in the x, y , z directions 
be u, v, w and U, V, W respectively. Suppose the collision is along the direction 
of «, U, so that v, w, and V, W, are unchanged. After collision u and U 
become u' and U'. 
For an elastic collision the relative velocity is reversed so that 

U'-u' = -(U-u). (4.43) 

44 Gases, liquids and solids 

The momentum is conserved, so that 

mu+MU= mi/+MU' 
or m[u-u']= -M[U-U1. (4.44) 

Combining equations (4.43) and (4.44) we have 

V = — !— [{M-m)U+2mu\. 

The gain in kinetic energy of the wall molecule as a result of the collision is 

1 M 

-Mil/'*- U 2 ) = j (t/'+ U\U'- U) 

2Mm . , 

■ [nut 1 - MU 2 + (M- m)uUl (4.45) 

(m+M) 2 

If we now consider the average behaviour of a large number of collisions 
we note that u must always be positive if collision is to occur, whereas U 
may be positive or negative and its average value must be zero. Consequently 
for a large number of collisions between gas molecules and wall molecules, 
the average value of uU is zero and the average gain of energy of each wall 
molecule becomes 

2Mm . —z — , 

<^Mr [mu MU] - <«*> 

where barred symbols refer to the mean square velocities of u and U. For 
thermal equilibrium there can be no net transfer of energy to the wall. Hence 

mu 2 = MU 2 . (4.47) 

Similar arguments may be applied to the other components (», V) and (w, W). 
The final conclusion is that for thermal equilibrium 

«? = MC 2 , (4.48) 

where c* and C 2 are the mean square resultant velocities of the gas and wall 
molecules respectively. This equation implies that for thermal equilibrium the 
colliding molecules must have the same average kinetic energy. 

4-4-2 Boyle's law 

For a given quantity of gas 

PV = f (total K.E. in volume V). 

Since the kinetic energy is constant if the temperature is constant this implies 

PV = constant. 

45 Perfect gases - bulk properties and simple theory 

4-4-3 Charles 11 law 

If the pressure is kept constant the volume increases linearly with the kinetic 
energy of the molecules. Only a sophisticated thermodynamic treatment can 
show that the kinetic energy itself is proportional to the absolute temperature. 

4*4-4 Avogadro's law 

Two different gases at the same temperature must have the same mean 
molecular kinetic energy, since they are each individually in equilibrium with 
the walls of their container. But 

PV = § (total K.E. in volume V) 

= \ (average K.E. per molecule x number of molecules in 
volume V). 

Hence (at constant temperature) PV will have the same value for all gases 
provided they contain the same number of molecules. 

4-4-5 Dal ton's law of partial pressure 

If we have two gases which separately have average molecular kinetic 
energies E t and E 2 , and a concentration of «i and n 2 per c.c. respectively 
we may write, 

P t = § (K.E. per unit volume) 

Pi = f (£!«l). 


Pi = f (E 2 m). 

For gases at equal temperatures E t = E 2 , so by adding the gases together a 
new concentration n = /ij + n 2 is obtained and the combined pressure is 

P = § En = §£(«i+n 2 ) = f «j£, + | n 2 E 2 

= Px+P 2 . (4.49) 

4-4-6 Mean free path 

We consider here the average distance between collisions of the gas molecules 
Suppose a is the effective diameter of a molecule, then any other molecule 
whose centre is within a distance a from the molecule considered will touch 
it [see figure 11(a)]. For our simplest model we assume all the other gas 
molecules to be instantaneously at rest and follow the fate of the single 
mobile molecule. If this molecule has, in a given instant of time, travelled a 
distance /, it will have swept out a volume na 2 l within which any other 

46 Gases, liquids and solids 


Figure 11. (a) The effective diameter a of a molecule. (b) A single molecule is 
allowed to travel through gas molecules considered instantaneously at rest. 

molecule will touch it [see figure 1 1(b)]. Thus if there are n molecules per c.c. 
the number of molecules with which our moving molecule will collide will 
be no 2 ln molecules. 
The mean distance between molecular collisions X may now be defined as 

(distance travelled) 
— ■, that is 

(number of collisions) 

X = —jt = — T • W-50) 

xa 2 ln no l n 

This is, of course, the simplest approach possible. In practice all the 
molecules are not immobilized. If we allow for the true relative velocity 
between the molecules this reduces k by a factor 1/V2. Again in a collision 
molecules will only rarely have equi-velocity head-on collisions and bounce 
away from one another; usually there will be some persistence of velocity in 
the original direction after collision and this will complicate the analysis. 
Maxwell also considered the influence of intermolecular forces. All these 
refinements merely change the value of A by a small numerical factor and we 
shall not consider this further. Some idea of the complexity of the problem 
may be gathered from Jeans* book on the kinetic theory of gases. 

A more important question is: how far does the concept of molecular 
collisions and molecular diameter affect our simple kinetic theory, where the 
basic assumption made was that the molecules have no size, and do not 
collide with one another in travelling across the container from one wall to 
the other? The answer is that the finite size of the molecules does influence 
the simple theory since the 'free' volume available for molecular movement 
is reduced. This will be discussed in greater detail when we consider the 
behaviour of 'imperfect* gases in Chapter 6. The collisions between the 
molecules do not. however, alter the simple theory. As we shall see in the next 
chapter collisions deflect molecules and remove them from the group under 
consideration, but these are constantly replaced (as a result of other collisions) 

47 Perfect gases - bulk properties and simple theory 

by molecules of other groups. Thus the behaviour is the same as though we 
could divide the molecules into non-colliding groups with specified velocities 
and directions. 

We see from equation (4.50) that X varies as 1/n, that is, it varies as the 
reciprocal of the pressure P. Taking an average value of a = 3 A., we find 
that for air at s.t.p. X a. 1000 A., whilst the mean distance between molecules 
is of the order (1/n)* a 30 A. These values 3, 30, 1000 A. for diameter, 
separation and mean free path are typical of gases under normal conditions 
of temperature and pressure. 

4-4-7 Softening of molecules 

Molecules are not hard impenetrable spheres. If the molecular speed increases 
they will, during collision, penetrate further into the repulsion fields of then- 
neighbours; but this effect is very small since the repulsive forces are very 

A more important effect is that which arises from the attractive or repulsive 
forces themselves. These will deflect the molecular paths and produce a 
marked change in the effective collision diameter of the molecules. An 
analysis for repulsive forces, falling off as the fifth power of the separation, 
was first given in analytical form by Maxwell. Forces of this type are not 
representative of most gas molecules. A more realistic model is one which 
considers the attractive forces which molecules exert on others passing 
nearby. This is illustrated schematically in figure 12. If the attractive forces 
are zero [figure 12(a)], molecule B will just hit A if the separation is less than 
a . If, however, attractive forces are not negligible B may well hit A even 
though its original direction is such that it would miss it [figures 12(6) (c)]. 

(a) (b) (c) 


Figure 12. The effect of attractive forces on the effective collision diameter. 

Hence although there is no change in the size of the molecule, from the point 
of view of collision and transport phenomena, it behaves as though it had a 
much larger effective cross-section a. Thus slow molecules will tend to collide 
even if they are very far apart, i.e. a tends to infinity as T tends to 0. On the 
other hand, for very high velocities, i.e. very high temperatures, there will be 
practically no deflection of the molecular paths and collision will only occur 
when the separation is close to a . A useful empirical relation for the effective 

48 Gases, liquids and solids 

molecular diameter a at temperature T in terms of the 'geometric' diameter 
ff is 

-*-*( ,+ D- 


where C is known as Sutherland's constant for the gas. Some typical values of 
Sutherland's constant are given in Table 11a, and Table 1 lb gives values of the 
effective cross-section of a number of gases at temperatures from - 100°C to 
300°C. The effect is not large but is not negligible especially for larger 

Table 11a Typical values of Sutherland's constant C 

Gas Ne He 2 Cl 2 Steam 

C°K. 56 72 125 350 650 

Table lib Effective cross-section compared with cross-section at 0°C. 

(Based on values of C given in G. W. C. Kaye and T. H. Laby (1959), Tables 
of Physical and Chemical Constants, 12th edn, Wiley). 

Gas -100° 0° 100° 200° 300° 













o 2 






Cl 2 






H 2 






4 , 4 , 8 Number of molecular collisions per unit area per second 

Consider a surface of area A subjected to molecular collisions. Divide the 
molecules into groups « t per ex. with velocity c t ; n 2 per c.c. with velocity 
c 2 , etc. Consider the first group n t and assume that they are all moving 
normal to the surface in question. Then in one second all the molecules in 
the volume aA will strike the surface, i.e. n^c x A molecules will strike. Hence 
the number of collisions per sq. cm. per second would be niCi, 

We now make use of the concept known as the Joule classification, which 
we used (without naming it) in our treatment of the kinetic theory. The velocity 

49 Perfect gases - bulk properties and simple theory 

c t of a molecule may be described in terms of its u, , »,, w u components. Since 
for any one group all directions are equally possible the number, on average, 
travelling in any one direction of given sign will be one sixth of the total 

Figure 1 3. Collision of molecules travelling with velocity d normal to a surface of 
area A. 

Hence the collision rate cm. -2 = \itiCi. 
For all the groups we have 

collisions cm. -2 sec. -1 = ^IniCt = \nc 

where n is the total number per c.c. and c is the mean velocity. 


4-5 Transport phenomena 

There are three bulk properties of gases which can be explained satisfactorily 
in terms of a simple molecular model, involving the transport of properties 
through the gas. For this reason they are called transport phenomena. 
Viscosity can be explained in terms of transport of momentum, heat con- 
duction in terms of transport of thermal energy, diffusion in terms of molecular 
transport produced by concentration gradients. 

4-5-1 Viscosity 

We consider the simplest arrangement where a plane surface YY is at rest 
and another parallel surface ZZ at a distance h away moves to the right 
with uniform velocity u. It is then found that a force F must be applied to 
the top plate to maintain the velocity u against the viscous drag of the 
fluid between the plates (figure 13). If the flow of the fluid between the 
surfaces is along lines parallel to the planes YY and ZZ (lamellar flow) it 
is found that F is proportional to the area A of the moving surface 
and the velocity gradient u/h. In the more general case where the velocity 
gradient is not uniform across the gap, F is proportional to A and to du/dy. 

50 Solids, liquids and gases 




area A 










Figure 14. (a) The concept of viscosity, (b) Model for describing transport of 
momentum in molecular terms. 

The factor of proportionality n is introduced such that 

Fsst >A~. (4.53) 

where n is the viscosity of the liquid or gas between the surfaces. 

We now show that the viscosity of a gas can be expressed in terms of 
momentum transported across the gas. Consider a plane XX within the gas 
parallel to the direction of the stream lines [figure 14(6)]. Let the velocity 
of flow at XX be u and the velocity gradient du/dy. We consider the molecules 
entering and leaving this plane. We assume 

(«) that the flow velocity u is very small compared with the mean gas 
velocity c; 

(6) that the only molecules reaching XX are those, which, on an average, 
have just made their last collision at a distance X from XX. Then at P 

the molecules have a flow velocity u+ — A and at Q a flow velocitv 


du , 

« X. 


From assumption (a) the number of molecules crossing an area A in one 

second is — A . From Q these molecules bring to XX horizontal momentum 

(du \ nc 
u X\ — A. 
dy J 6 

From P a similar flow brings horizontal momentum 

/ du \ nc 
m [ U+ dy A J-6 A - 



51 Perfect gases - bulk properties and simple theory 

tic A 

The plane XX itself is also discharging molecules per second towards 


both P and Q. The resultant behaviour is summarized in Table 12. 
Table 12 Molecules entering and leaving an area A in plane XX per sec. 

Number Horizontal momentum 

Entering from P 


Entering from Q 


Leaving from XX upwards 


Leaving from XX downwards 



\ dy ) 6 
\ dy } 6 








Net addition to XX 

We see that there is no accumulation of molecules or momentum in plane 
XX. There is, however, a transport of momentum per second through the 
plane given by the difference between equations (4.55) and (4.54). This 
amounts to 



mine , du 
= -T A ay- <**» 

This horizontal momentum per second is a horizontal force which is trans- 
mitted through the gas to the top moving surface. As we saw before, in 
equation (4.53), the tangential force on the top plate is 

x, , du 



By comparison we see that we have derived a relation for the viscosity of the 

n = \mXnc (4.56a) 

i m - 
or 1 = 3 — i c (4.56b) 

52 Gases, liquids and solids 

We conclude that the viscosity of a gas is proportional to £, that is to 7** and 
is independent of n, that is, of the pressure. Both these conclusions are well 
supported by experiment. If however the pressure is too low and the mean free 
path becomes comparable with the distance between the moving surfaces, 
equations (4.56) are no longer reliable. This is discussed in the following 
paragraphs, which may be omitted on a first reading. 

( , C) du u 


Figure 1 4. (c) Ignoring the problem of the solid boundaries the velocity gradient 
across the bulk of the gas is -. (rf) If impinging molecules from the last mean- 
free-path separation acquire the tangential velocity of the solid wall there is an 
effective discontinuity in the mean tangential gas velocity close to the wall. This 
leads to a reduction in the velocity gradient in the bulk of the gas. (e) If impinging 
molecules rebound 'elastically* so that their horizontal velocity component after 
impact is the same as before, no horizontal momentum is communicated to the 
walls. There is complete slipping of the molecules over the walls and zero velocity 
gradient in the bulk of the gas. 

Suppose the distance between the plates is y, the lower plate is stationary 
and the upper plate has velocity «. If v is very large compared with the mean 
free path X, the velocity gradient across the gas [see figure 14(c)] is 

du _u 
dy y 


Let us now consider in greater detail the situation at the solid plates them- 
selves. It is clear from our derivation of tj that at the solid walls only one 
half of the transport process is operative since molecules cannot pass through 
the walls. Some change in flow rates must occur close to the walls if the 
appropriate shear stress is to be communicated through the gas to the 
solid wall. 

Consider first the stationary wall. Molecules arriving after their last collision 
at a mean distance X away, bring with them horizontal velocity «,. If all the 
molecules striking the wall are absorbed and then re-emitted with the 
tangential velocity of the (stationary) wall, this tangential velocity u 2 is zero 

53 Perfect gases - bulk properties and simple theory 

[see figure 14(d)]. Thus the average velocity of the molecules very close to 
the stationary wall is the average of «i and zero, that is uJ2. This is equivalent 
to saying that the molecules slip over the walls with velocity uJ2. A similar 
slip occurs at the moving surface. Between these two planes at which slip 
occurs there is full transport of molecules to and from the plates, so that a 
constant velocity gradient may be maintained throughout the bulk of the gas 
in accordance with equation (4.57a). The velocity gradient is determined by 
the fact that between the stationary surface and the first mean free path there 
is a change in velocity of uJ2 over a distance A, 

thatis * a* 
2 dy 

The velocity difference v between the gas near the stationary wall and the 
gas near the moving wall [see figure 14(A)] is clearly 

2u i 


v = U—2A-. 

Over this region there is a uniform velocity gradient of amount 


v _ \ dyj 


y y 


du u 
dy y+2X' 


We now see that the shear stress across every section of the bulk of the gas 
is communicated to the solid walls by the velocity jump close to the solid 
surface. Comparing equation (4.57b) with (4.57a), the apparent viscosity 
of the gas is reduced by a factor 



Clearly if the mean free path is an appreciable fraction of the separation there 
will be a large reduction in the apparent viscosity. In the limit if A is greater 
than y, none of the concepts used in this analysis is applicable. 

There is yet a further factor involved. The molecules striking the surface 
may not all take up the tangential velocity of the surfaces before rebounding. 
In the extreme case, if the tangential velocities of the molecules leaving the 
surface are the same as those striking it, there will be no transfer of tangential 
momentum to the walls. Consequently there will be no resultant velocity 

54 Gases, liquids and solids 

gradient in the bulk of the gas, but a velocity discontinuity of «/2 at a short 
distance from each of the walls [figure 14(e)]. This corresponds to perfect 
'slipping' of the gas over the wall surfaces without the transfer of any shear 
stress. Under these conditions the gas would appear to have zero viscosity. 
In practice the situation is much closer to that involving adsorption and 
re-emission with the wall velocity. As a result the simple correction, given by 
equation (4.57c), is valid. 

Finally we may refer the reader to a more quantitative treatment of this 
problem given by S. Chapman and T. G. Cowling in The Mathematical Theory 
of Non-Uniform Gases, Cambridge University Press, 1939, chapter 6.* If is 
the fraction of molecules absorbed and emitted with the tangential velocity 
of the wall, then the fraction reflected with the same tangential velocity as 
the incident molecules is 1-0. The average tangential velocity of the 
'reflected' molecules, for example from the stationary wall, is 

u 2 = O-0)u. (4.57d) 

The mean velocity at the wall is i(«i+« 2 ). This is the velocity of slip. 

«i - i(«i + «*) * A — . (4.57e) 


Using equation (14.57d) this gives for the slip velocity 

,. , . 2-0 , du 

¥.u l + u 2 ) = — A — . (4.57f) 

Because of similar slip at the moving wall the velocity difference across the 
bulk of the gas is 


X J y ' «.57g) 

The apparent viscosity of the gas is thus reduced by the factor 




For = 1 this reduces to equation (4.57c); for = 0, the ratio is infinite. 
This corresponds to complete slip at the walls as shown in figure 14(c). 

4 -5 -2 Thermal conductivity 

Consider heat flow across the material lying between parallel surfaces YY 
and ZZ [figure 15(a)]. If the temperature of the hotter surface is T t and the 

* I am indebted to Mr N. J. Holloway and Mr B. Scruton for bringing this to my 

55 Perfect gases - bulk properties and simple theory 






' dy A 

Figure 1 5. (a) The concept of thermal conductivity, (b) Model for describing 
transport of thermal energy in molecular terms. 

T —T 

colder T 2 , the rate of flow of heat is proportional to — and to the area 


of the surfaces. More generally we write for the heat flow per second 



where K is the thermal conductivity of the material and dT/dy the temperature 
gradient. The negative sign is because there is a flow of heat from the hotter 
to the colder region (positive Q for negative dT\dy). 

We again assume that in a gas heat conduction is due to the transport 
of thermal energy. If c v is the specific heat at constant volume per molecule 
the rate of transport of thermal energy into plane XX from planes at a distance 
X from it is: 

from the upper plane: 


c v 


from the lower plane: 

nSA /, dT,\ 
T C " \ T -Jy X ) 



Again we may show that there is no net accumulation of molecules or of 
thermal energy in plane XX. There is, however, a net transport of thermal 
energy per second of amount equal to the difference between equations 
(4.59a) and (4.59b). 

_ ncX .dT 

Q = — CyXA— . 

3 dy 

56 Gases, liquids and solids 

By comparison with equation (4.58) we see that the thermal conductivity is 

given by 

_ ncl 

K= — Cy. (4.60) 

As in the preceding discussion of viscosity the temperature gradient near 
solid walls will depend on the degree of thermal equilibriation between the 
temperature of the molecules striking the walls, and the walls themselves. If 
the molecules acquire the temperature of the walls, the temperature very close 
to the wall will show a small discontinuity, and there will be a reduction in 
the temperature gradient through the bulk of the gas. If the collisions are 
perfectly elastic so that the molecules strike the wall and are reflected with 
their original thermal velocity, that is they ignore the temperature of the wall, 
no heat can be transferred to or from the walls. There will be a large dis- 
continuity in temperature at the walls: the total temperature drop will occur 
in a very short distance comparable with A and there will be a negligible 
temperature gradient across the mass of the gas. Consequently there will be 
no heat transfer. This is the limiting case in which the 'heat transfer 
coefficient' is zero. 

Ratio of thermal conductivity and viscosity. From equations (4.60) and 
(4.56a) we have 

K c y c r N C, 


V m mN M 

where C r is the specific heat per gram-molecule and M the molecular weight. 
We therefore expect to find 

KM , 

7cv = h (4 - 61 > 

The experimental value of this ratio lies between 1-4 and 2-5 for a very 
wide range of gases. The discrepancy is due partly to molecular repulsions 
which tend to reduce n. A more important factor is that, in thermal conduction, 
the transport of the translation^ kinetic energy takes place more efficiently 
than the transfer of the rotational and vibrational forms of thermal energy. 
This leads to an increase in K relative to n, and more sophisticated theories 

are available that agree well with the experimental ratio of . Even the 

order of agreement with the simple theory provides strong support for the 
essential validity of the transport mechanisms described above. 

4-5-3 Self diffusion 

If we have a concentration gradient in a gas, the molecules will move from 
the more concentrated to the less concentrated regions via molecular collisions ; 

57 Perfect gases - bulk properties and simple theory 

this involves a kinetic diffusion process. If over a distance dx the concen- 
tration increases by dn molecules per c.c. the concentration gradient is dn/dx. 
It is then found that the number of molecules per second crossing an area 
A normal to the gradient, i.e. the rate of diffusion of molecules per second, can 
be written 

dN n dn , 

— = -D—A, 
dt dx 

where D is called the coefficient of self diffusion and the negative sign implies 
flow in the direction of smaller concentration. 
Consider again a plane XX where the concentration is n and neighbouring 

planes P and Q distant from it where the concentrations are n-\ — and 


n — - respectively. The number of molecules per second crossing from P 

1 / dn\ 
= -\n + — A)< 
6\ dx ) 

I cA (4.62a) 

i \ ax / 

and from Q 

/ //« \ 

i cA (4.62b) 

-H'-s 1 )' 

There are also molecules leaving on each side of XX of amount \ncA. Again 
there is no net accumulation of molecules in XX, but the net transfer is 

kc dn . n dn , 

A = -D—A. 

3 dx dx 


D = * 

It is interesting to see the orders of magnitude involved in D. For air at 
s.t.p. we have: 

a = 2-5 A.; A = 1000 A.; c = 45000cm. sec. -1 

This gives a value of D of order of 1 molecule per sq. cm. per unit concentration 
gradient. D is very small compared with c due to the retarding effect of 
molecular collisions. 

There is only one direct way of studying D. This is to incorporate some 
radioactive molecules in one specimen of gas and place it in contact with an 
identical gas containing no radioactive molecules. The diffusion of the 

58 Gases, liquids and solids 

radioactive species may then be followed, say, using a Geiger counter. Even 
here there is some difficulty since the radioactive species may not be the same 
size as the non-radioactive. For this reason most diffusion experiments 
involve the diffusion of one gas through another. If the concentration of gas 1 
is «! per c.c. and that of gas 2 is n 2 per c.c; if c x and c 2 are the respective 
molecular velocities and ay and a 2 the respective molecular cross-sections, 
the diffusion coefficient of gas 1 into gas 2 may be shown to be given by 

A,2 = 

Vicl + cl) 

'iit{n 1 J rni\a 1 -\-a 1 y 1 


A typical example of D 1>z is the diffusion of a vapour through air. Consider 
a cylindrical vessel of cross-sectional area A (figure 16). It contains a small 

n = 

cold surface 

saturated vapour 


Figure 1 6. Diffusion of molecules from the surface of a liquid where vapour con- 
centration corresponds to that of saturated vapour. The vapour condenses on a 
cold surface at the top of the vessel so that at this region the vapour concentration 
is virtually zero. 

quantity of liquid at the bottom, and at the top open end is a cold surface on 
which the vapour immediately condenses. The vessel is full of air and diffusing 
vapour. Just above the liquid the vapour is at its saturation pressure implying 
a concentration n molecules per c.c. At the cold surface the vapour pressure is 
almost zero. In the steady state there can be no accumulation of vapour 
molecules in any section. This implies that the diffusion rate across every 
section is a constant. Consequently there must be a linear concentration drop 
from n at the surface of the liquid to zero at the cold surface, distance h. 

m dn n 
Then — = _. 
dx h 

Hence the mass of vapour transferred per second and condensed on the cold 
surface is 

_ . dn _. . « 

w = D U 2 A — m= Z> 1>a A -m, 
ax h 

59 Perfect gases - bulk properties and simple theory 

where m is the weight of the molecule. The product nm is the density p of the 
saturated vapour. If we, therefore, determine the weight w of liquid condensed 
on the cold surface (or lost from the liquid) per second, we have 

_ wh 

A,a = — . (4.65) 


In this way it is possible to deduce values of (oi +a 2 ) 2 . Some typical values 
of a deduced from measurements of n, k and D are given in Table 13. 

Table 13 Molecular diameters a (10~ 8 cm.) 


Deduced from 


K D 



3-2 3-8 

co 2 


3-8 4-5 



3-7 3-8 

Cl 2 


5-9 — 



1-9 — 

H 2 


2-7 2-7 



4-3 40 

N 2 


3-6 3-7 

o 2 


3-5 3-6 

(Adapted from G. W. C. Kaye and T. H. Laby (1959), Tables of Physical and 
Chemical Constants, 12th edn, Wiley.) 

4-6 Sound waves in a gas 

4-6*l Velocity of sound in a gas 

All elastic media can transmit waves of well-defined velocity and frequency 
and with solids the waves can involve shear as well as compression. But with 
gases the only waves that can be transmitted are those involving a succession 
of compressions and rarefactions. The velocity of such a longitudinal wave 
is given by 


elastic modulus .. ,^ 

- A — zr • (4.66) 


The elastic modulus is the bulk modulus defined as the pressure increment dP 

60 Gases, liquids and solids 

divided by the fractional volume increment dVjV. Since a pressure increase 
produces a diminution in volume it is usual to define the modulus as 



We shall soon show that sound waves are adiabatic and not isothermal. 
However, at this point, we shall assume isothermal waves. If we write 

PV = RT 

then for an isothermal change PdV+ VdP = RdT = 0, so that the modulus 


= P. (4.67) 

Hence from equation (4.66) 

IP Inmc* /c*\* 

since the density p = run. 

» = («*)* (4.69) 

where u* is the mean square molecular velocity in one direction. The value of 
(« 2 )* is very nearly equal to u; this implies that the velocity of a sound wave 
is approximately equal to the mean molecular velocity in the direction of 
propagation. This at once tells us that the molecule is the messenger which 
carries the impulse through the gas. In this way we have linked the equation 
which describes wave propagation in terms of bulk properties [equation 
(4.66)], with a molecular process. 
For an adiabatic wave the gas equation is 

PV = constant. 
This leads to a value of the modulus 

= yP 



v p 


Consequently the adiabatic wave velocity is\?y times that given in equation 
(4.68). This increases v by about 20 per cent for sound in air, but does not 

61 Perfect gases - bulk properties and simple theory 

change the basic conclusion that sound waves are propagated by molecules 
communicating an impulse to neighbouring molecules by collision with them. 

4-6-2 Why sound waves are adiabatic 

It has long been a tradition in text books of sound to assert that in practice 
sound waves are adiabatic, and not isothermal, because their frequencies are 
so high. There is a temperature rise at the compression and a cooling at the 
rarefaction, and this implies that the time taken for the wave to pass along 
(so reversing the position of compressions and rarefaction) is not sufficient for 
thermal equilibrium to be reached. Consequently the wave has not the time to 
become isothermal. This is true but we shall now show that this is because the 
frequency is relatively low and that high frequencies (in theory) would favour 
isothermal conditions. Basically this is because the time needed for temperature 
equilibriation to occur turns out to be proportional to the square of the 
wavelength, whereas the time available for this to occur is simply propor- 
tional to the wavelength. Thus equilibriation occurs only if the wavelength / 
is less than some critical length l c as indicated schematically in figure 17(a). 



time needed if 
thermal equilibrium 
is to occur 

time available for 



wavelength / 

L — i — J< — i — J 

compression rarefaction compression 

Figure 17. (a) Time factors involved in achieving isothermal conditions in a 
sound wave ; these are attainable only if the wavelength is less than lc. (b) The 
compressions and rarefactions in a sound wave. 

The following simple analysis shows how this arises: 

Consider an instantaneous snapshot of a plane wave travelling from left 
to right with velocity v, frequency v and wavelength /. There are compressions 
at A, C, and a rarefaction at B [see figure 17(6)]. Suppose the heating at A 
and the cooling at B (under perfect adiabatic conditions) give a temperature 
difference between A and B of T. The temperature gradient is of course 
sinusoidal but for simplicity we shall assume that it is linear. Then we may 



T _ 




62 Gases, liquids and solids 

As a result heat flows from A to B tending to equalize the temperature and so 

reducing the temperature gradient. As we are only interested in an order of 

magnitude calculation we shall ignore such changes and assume that the 

gradient remains at — — . 

If K is the thermal conductivity of the gas the rate of heat flow AQ per 
sq. cm. of section in time Ar is 


— = -K— = K— . (4.72) 

A/ dx I 

Consequently the time A/ taken for a quantity of heat AQ to flow is 

A/ = ^. (4.73) 


Equalization of temperature between A and B demands that the heat AQ 
raises the temperature of the mass of gas between A and B by T, that is, we 
need an amount of heat, 

AQ = volume x number of molecules per c.c. x specific heat per 
molecule x T 

= ~nc P T. (4.74) 

(For an instantaneous snapshot of the wave, c r is the appropriate specific heat.) 
Substituting this in equation (4.73) we obtain 

*-!F- (4.75) 

Now the time available for this equilibrium to occur cannot be greater than 

the time taken for the compression and rarefaction to change places, i.e. 

about t/2 where r is the period of the wave. A more realistic estimate is 

r I 

— = — . Consequently temperature equilibrium will occur, i.e. we shall 

4 4t> 

approximate to isothermal conditions if 

— > 




— > 


l*nc r 







v > 

ncpV 2 

itCpV K 

In a previous section we showed [see equation (4.60)] that for a perfect gas 

K=* — (c K ), 
63 Perfect gases - bulk properties and simple theory 

where A is the mean free path. Hence isothermal conditions will be achieved 
if the frequency 

3nc P v 2 . 3v 2 

* > —r. — . ie. v > — (4.77) 

ncXcr cX 

ignoring the difference between c P and c v . 

Since v is of the order of c s: 3 x 10* cm. sec. -1 and A is about 1000 A. 
(10 - 5 cm.), this requires a frequency greater than 10 9 cycles sec. -1 . This is 
extremely high. Further if we do the same substitution for / we find that / 

must be less than —A, that is, it must be less than the mean free path in the 

gas. Under these conditions a 'normal' sound wave cannot be propagated. 
We conclude that sound waves are adiabatic because their frequency is not 
high enough for them to become isothermal. In theory they could become 
isothermal at frequencies of the order of 10 lo sec. -1 but such frequencies 
could not be transmitted through a gas as a normal sound wave. 

64 Gases, liquids and solids 

Chapter 5 

Further theory of perfect gases 

In this chapter we shall first describe a rather more sophisticated kinetic 
theory of perfect gases. Part of the exercise is purely computational and, 
although it looks more impressive, adds little to our physical understanding. 
There are, however, a number of points which emerge which are interesting 
and useful and which shed new light on some of the assumptions made in the 
simpler forms of the kinetic theory. We shall also discuss the velocity dis- 
tribution in a gas and the thermal energy of its molecules. 

5-1 A better kinetic theory 

5-1 -1 Assumptions 

First we recapitulate our basic assumptions. Let us assume that we are dealing 
with a very large number of molecules uniformly distributed in density; 
that they have complete randomness of direction and velocity; that the 
collisions are perfectly elastic; that there are no intermolecular forces; and 
finally that the molecules have zero volume. 

We now consider a way of describing their distribution in space. Thus to 
each molecule we attach a vector representing its velocity in magnitude and 
direction [figure 18(a)]. We then transfer these vectors {not the molecules) to 

(a) (b) 

radius r 

Figure 18. (a) Velocity vectors are attached to the individual molecules, (b) The 
velocity vectors are all transferred to a common origin. A sphere of arbitrary radius 
r is drawn about the origin. 

a cotamon origin [figure 18(*)] and construct a sphere of arbitrary radius r, 
allowing the vectors to cut the sphere (if necessary by extending their length). 

65 Further theory of perfect gases 

Then the velocity vectors intersect the sphere in as many points as there are 

If we postulate randomness of molecular motion all directions are equally 
probable, so that these points will be uniformly distributed over the surface 
of the sphere. Suppose we consider AN molecules, where N is the total 
number present. These could be the total number of molecules in the vessel 
or the number per c.c. or the number per c.c. with a specified velocity range. 
Then the number of vector points corresponding to an element of area AA 
on the sphere will be 

AN = ^;M 
4/tr 2 


We can specify the element AA in terms of spherical co-ordinates and <l> as 
shown in figure 19. 




Figure 1 9. Specification of an element of area on the velocity sphere (a) in terms 
of angles 6 and 4, (b) in terms only of angle 9. 

AA = rdOxr sin d$. 


Thus the number of molecules travelling in a direction between and 
0+dO and between ^ and #+d<j> (relative to some arbitrary axes) is 


AN. t = -N =—xsindd0 d<f>. 

Anr 2 4n 


We see that r disappears from the result: AN depends only on the specified 

We could continue to use equation (5.3) in our subsequent calculations, 
but this would make the arithmetic more complicated than is necessary. In 
all the cases we are interested in, we only need to consider molecules travelling 
in a direction between 6 and 0+dO irrespective of their ^ position. That is to 

66 Gases, liquids and solids 

say it is sufficient for our purpose to make AA an annulus on the sphere 
lying between angles and 0+d0 [figure 19(A)]. 

Then AA = Inr sin 9 rdO, (5.4) 

so that the total number of molecules travelling in this direction is 

AN, = -=?- N = i sin d0N. (5.5) 


5 • 1 • 2 Number of collisions with a solid wall 

We use the above result to calculate the number of molecules colliding per 
second with a square centimetre of a solid surface. Let us first divide the 
molecules into groups according to their velocities. Suppose there are n c 
molecules per c.c. with velocity between c and c+dc. Then the number out 
of this group that at any instant are travelling towards the surface from a 
direction 0, 6 +d6 is 

A» c , i = \ sin dO x n c . (5.6) 

The only molecules that hit the surface of area a in time dt are those contained 
within a prism of basal area a and of sloping height cdt [see figure 20(a)]. The 


Figure 20. (a) The number of molecules hitting the area « in time dt is equal to the 
number of molecules within the prism of basal area a and sloping height cdt. 

volume of this is cdt a cos 0. Hence the number of molecules in this group 
hitting a in time dt is 

Ak c ,,x cdt a cos 0. (5.7) 

The number striking each sq. cm. per sec. is this number divided by ccdt 

= i sin cos 0d0 n c c. (5.8) 

For all values of from to w/2, i.e., for the whole of the 'half-space' above 
the surface the total collision rate is 


\ sin cos d0 n c c = fac. (5.9) 


67 Further theory of perfect gases 

For all molecules of all velocity groups the number striking each sq. cm. 
per sec. is then 

i I*cC = \nd, (5.10) 

where n is the total number per c.c. and c is the mean velocity defined by 

c = . This result differs from that given in the previous chapter 


[equation (4.52)] where the fraction in front of nc was found to be £ instead 


5-1 -3 Gas equation of state 

We now consider the momentum transfer per second per sq. cm. of wall, 
i.e., the pressure exerted by the gas. We first consider the group of molecules 
« e per c.c. whose velocity is between c and c+ dc. Of these consider the group 
approaching the surface at a direction 6,6+dd; each molecule brings with it 
momentum mc. The vertical component is, therefore, mc cos 6 and after 
collision it is reversed so that the net momentum transferred to the wall is 
ImccosO. The horizontal component mc sin 6 is unchanged so that no 
tangential momentum is communicated to the wall. The vertical momentum 
change per second transferred to each sq. cm. of wall, i.e. the pressure, is 

2mc cos x (number hitting each sq. cm. per sec.) 

= 2mc cos i sin cos dO n^c 

= nuicC 2 cos 2 sin dO. (5.11) 

Integrating over the whole half-space we obtain the pressure produced by this 
group of molecules: 

mice* 1 cos 2 sin dd 

J O 

-«u>[ ros ; e ] ! =w*. 

For all groups of molecules we thus obtain 

p = imnc*, 

where the mean square velocity c 2 is given b 

2/icC 2 




5 • 1 -4 Transport phenomena 

Consider for example the viscosity of a gas in terms of the momentum 
transport across a plane as described in the previous chapter. The flow is 

68 Gases, liquids and solids 

assumed streamlined parallel to the XX plane, the velocity gradient du/dy 
in the steady state is constant. We have to calculate the transport of horizontal 
momentum across each sq. cm. in the plane XX per sec. Consider a small area 
element da in XX. The only molecules we are interested in are those which on 
an average made their last collision at a distance X from the element. Draw 
a hemisphere of radius X with the element as centre [see figure 20(ft)]. Those 

Figure 20. (b) Transport of momentum through element of area doc. A hemisphere 
of radius X is constructed with da as centre. All molecules from this hemisphere 
have, on average, just made their last collision before proceeding towards da., from 
half of the gas. 

coming from the direction 6,6+ dO, are at a distance X cos 6 from XX so that 
these molecules bring additional horizontal momentum mX cos 6 (—) . 

The number coming from this direction and striking each sq. cm of the ele- 
ment per sec. is given by equation (5.8). The transfer of horizontal momentum 
per sq. cm. per sec. is therefore: 

«* «* * ( j- ) * i sin 6 cos (5 . M) 

If we integrate this over the whole of the top half-space we obtain 



There is a similar contribution from the half-space below so that the tangential 
stress in the plane XX is simply 




If we equate this to the viscous stress // — we obtain 

1 = inrnsX, (5 17) 

for molecules of this group. For all groups of molecules 

69 Further theory of perfect gases 

The result is identical with that obtained from the Joule classification. 

We may explain this result in a simple way. The number of molecules 
striking each sq. cm. in XX per sec. is \nc instead of \nc as in the Joule 
derivation, but whereas in the Joule derivation each molecule is assumed to 
come from a constant distance X from plane XX, in the present treatment the 
last collision occurs on the hemisphere of radius X, so that the last collision 
distance measured normal to XX ranges from to X. The average effective 
distance is f/l. The horizontal momentum transfer is thus the same in both 

5*1-5 Mean free path X and collision frequency v 

We now extend our treatment to allow for the finite diameter a of the mole- 
cules. As we saw in the previous chapter, a very simple result for the 
average distance X between each collision is given by 

X = Ar- (5.W) 


There are many sophisticated ways of improving this relation but they only 
change A by a small numerical factor. 

We now define the collision frequency v as the average number of collisions 
per second made by each molecule. 

Average distance travelled per sec. 
Average distance between collisions 

v = 

v = 

c (5.20) 

This result does not depend on the way in which X is derived. 

5 • 1 -6 Distribution of free paths: survival equation 

We wish to follow the fortunes of a group of N molecules. As they travel 
they collide with themselves and other molecules. Can we estimate the number 
that, at any specified stage, have not yet made a collision? 

Suppose that, at some instant, n have survived without collision. If each 
is allowed to travel a further distance dx along its free path further collisions 
may occur. We assume that this number of collisions is proportional to both 
n and dx. Hence the number removed by these collisions is proportional to 
ndx. To show that n is decreased we write: 

dn = -Pndx, (5.21) 

where jP is the collision probability for the gas. We assume this to be a constant 
for a given gas under specified conditions. Hence, 

70 Gases, liquids and solids 


= -Pdx 


n = Nc~ rx 
dn = -PNa- px dx. 



n c (number having collided) 


distance travelled 

Figure 21. Graph showing survival of an initial group of N molecules. After 
travelling a distance of x. n have survived without collision whilst the number 
rt c suffering collision is given by nc = N—n. 

Since n is the number surviving after travelling a distance x, the number n c 
which have suffered collision (see figure 21) is simply 



n c — N—n. 

dn c = d{N-n) = -dn, 

dn c = +PNe- Px dx. 

This is the number of molecules with mean free path between x and x+dx. 
The average mean free path x is given by 

X = 

xdn e 


71 Further theory of perfect gases 


i.e., X = ±«- 

xPNe~ r 'dx 



Consequently, equation (5.22) becomes 

n = JVe K 


This is the basic 'survival equation'. » is the number of molecules out of a 
group N that have not yet made a collision after travelling a distance x. 

5*1'7 Collisions with surfaces: a final treatment 

We may now ask a simple but important question in relation to the kinetic 
theory as we have so far developed it. We have always assumed that if some- 
where there is a group of molecules travelling in some specified direction we 
can calculate the number striking a sq. cm. of a surface per second by com- 
pletely ignoring intermolecular collisions. The previous section shows that 
this is quite unjustified. Molecules a long way from the surface never get there : 
they are diverted by collisions on the way with themselves and other molecules. 
Consider at any instant an element of volume dV in a gas of uniform 
density containing n molecules per c.c [figure 22(a)]. The number in the 



Figure 22. (a) Molecules in element of volume dV, after their last collision, start 
travelling towards AA (A) Conical annulus of volume 2w sin rdO dr defines dV. 

element is ndV. 

If v is the collision frequency, the number of collisions occurring within 
dV in time dt is 



72 Gases, liquids and solids 

The i is introduced since otherwise each collision would be counted twice. 
Each collision starts off two new free paths so the total number of free paths 
originating in time dt from the element dVi& 


Because all directions of molecular motion are equally probable in any 
element of the gas, these molecules start off uniformly in all directions. The 

fraction heading towards the area AA is — , where Aco is the solid angle sub- 

tended by AA at the volume element. 

. AAcosO 
Aco = . 

The number heading towards AA is thus 

An = —xvndVxdt. (5.27) 

However, from the survival equation, the number of molecules reaching AA 
without having made a collision, and so having been eliminated is 

_ r 

Arti = AnoK 

Consequently, the number of molecules leaving dV in time dt and ultimately 
reaching AA without having made an intermediate collision is 

. 1 AA cos 6 - - .„ „ 

to — 7 1 — V " e dVdt - (5.28) 

We note that An,, depends on the time element dt, not on the absolute time. 
This is because on the average any given element of volume appears un- 
changed. (To some extent this observation, in itself, implies the conclusion we 
shall draw in the following paragraphs.) We may thus choose our time element 
dt for each volume element dVaX various times, but in such a way that all the 
undeflected molecules from all the volume elements arrive at AA during the 
same time interval dt. This enables us to integrate for the whole volume of 
gas above AA. Once again we choose as our element of volume a conical 
annulus of radius r sin 0, thickness rdB, height dr [figure 22(A)], so that 

dV — 2mr sin 6 rdB dr. (5.29) 

We substitute in equation (5.28) and integrate for r = to r = w, and for 

= Oto0»-. This gives the total number of molecules, from all half- 

73 Further theory of perfect gases 

space, from all directions striking AA in time dt as : 



' J o 

= ivnX LA dt 

2 cos sin 

d9e *dr 

The number striking per sq. cm. per sec. is then 

But, irrespective of our model for calculating X, 




v = 

Hence the number striking each sq. cm. per sec. is 

inc. (5.32) 

This is the same as the result we have already obtained. It implies that 
although molecules are continuously colliding and changing their directions 
and velocities, other molecules are replacing the ones that are eliminated. 
For this reason the simple picture which ignores intermolecular collisions in 
calculating wall collisions, gas pressure and transport phenomena yields 
the correct answer. 


Sedimentation under gravity 

Consider a gas column of unit cross-section and at uniform temperature 
T (figure 23). We shall show that because of gravity there is a density (and 

B n + dn 



A n 


Figure 23. Decrease of density and pressure with height of an ideal gas. 

pressure) gradient in the gas. Our increments are measured positively in the 
direction of increasing height h. 

74 Gases, liquids and solids 

At the level A, height h, let the pressure be p and the molecular density n 
per c.c. At level B, height h + dh, the pressure is p + dp and the density n + dn. 

The resultant force on the layer of gas between A and B is a pressure dp 
downwards and a gravitational force nxdhxmg downwards. For equilibrium 

ndhmg+dp = 0. (5.33) 

Since p = \nmc J - and &■ is constant at all levels (because of constant tempera- 

dp = $me'dn. 
Equation (5.33) becomes 

n imc 2 
Integrating from h = to h and from n to n we obtain 


$mc 2 \Nmc 2 

, « , p Me , 

«o Po «r 

For air, substituting M ~ 30, i{ = 8-4x 10 7 , T= 300, we find that at a 
height of 1000 ft. the pressure is reduced by about one thirtieth of an atmo- 

5 -2 -2 Sedimentation of particles 

If particles of mass m and density p axe suspended in a liquid of density 
p , then if there is complete thermal equilibrium, the concentration of particles 
is given by 

ln « = _^( £ - £2 ) 
«o RT p 

An examination of the variation of n with height for a fine suspension was 
used by J. Perrin in 1908 to calculate JV . He obtained a value of 6x 10", 
which is virtually identical with the value of Avogadro's number obtained 
by other methods. This shows that fine particles in thermal equilibrium behave 
like a gas of molecular mass m. It is quite wrong to imagine that this is simply 
because they are buffeted around by the molecules of the liquid in which they 
are suspended. They possess this property simply by virtue of being in thermal 
equilibrium with themselves and their surroundings. Of course the collisions 
between the particles and the liquid molecules ensure that the whole system 
is in thermal equilibrium, but the kinetic theory applies to the particles 
themselves just as it does to the molecules of a conventional gas. Equation 

75 Further theory of perfect gases 

(5.35) also shows that by increasing g the major fraction of the particles will 
be formed in a thin layer near the bottom of the vessel. This is the basis of the 
action of a centrifuge. 

5*2'3 The Boltzmann distribution 

We may rewrite equation (534) in the form 

« RT kT 

where k = — = the Boltzmann constant. 


Then « = «o exp ( -'^A I . (536) 


Since the temperature is constant throughout the gas the kinetic energy is 
constant at all levels; only the potential energy has been changed, by amount 
mgh. Then mgh is the amount by which the energy of a molecule at h exceeds 
that at ground level; call this e. Then 

n a Roexpf —— | . (537) 


exp ( —r 1 ],. (538) 

Boltzmann showed that for any equilibrium distribution of molecules 
this relation generally holds. We may write 

(number of molecules at energy _ (number of molecules / e_\ 

level e above ground level) ~ at ground level) xp ^ /fcr/' 

Again for two energy states c t and e 2 

n t CXP (-|?) (_Ae\ 

where Ae is the energy gap between states 1 and 2. The Boltzmann function 
thus describes the relative population of energy states in an equilibrium 

5*3 Temperature variation of reaction rates 

A chemical reaction can only occur spontaneously if the final system has 
a lower free energy state than the initial system. In general however, before 
the reaction can take place one or more of the reactants must be excited to a 
higher energy state (see figure 24). The rate of forward reaction is then 

78 Gases, liquids and solids 




Figure 24. Forward reaction has an activation energy if. the backward reaction 
an activation energy e$. The heat of reaction is q = tb- ef. 

determined by the number of molecules which have enough energy to 
get over this barrier. The forward reaction rate is proportional to 

exp ( — — } where e f is the activation energy. 

It is interesting to note that the back reaction rate will be proportional to 

exp | — — I, and that the heat of reaction is equal to the difference between 

\ kT ) 
the activation energies of forward and backward reaction, 

q - e t -e f . 

5-4 Distribution of velocities in a perfect gas 

5-4-1 Velocity distribution in a one-, two- and three-dimensional gas 

We expect to find a wide distribution of molecular velocities in a gas, because 
even if we could start off with all the molecules travelling in different directions 
with equal velocities, random collisions are bound to speed some and retard 
others. Consequently we expect to find the velocities ranging from to oo* 
but grouped around a definite average value determined by the temperature. 
The velocity distribution is, in fact, of the same type as the Boltzmann 

If there are no intermolecular forces the only energy the individual molecule 
has is its kinetic energy \mc*. (We ignore here vibrational or rotational 
energies which are assumed to be unchanged by the gas-collision process.) 

* Because of relativity effects an infinite velocity would imply an infinite effective mass. 
However, the fraction of molecules with velocities high enough to produce appreciable 
relativity effects is negligible. In what follows we shall assume that relativity effects may 
be ignored and that the velocity distribution extends to infinity although, of course, 
infinite velocities cannot in fact occur. 

77 Further theory of perfect gases 

Then according to the Boltzmann distribution the population density of 
molecules with velocity c is proportional to 


\ 2kT) 


If we attach velocity vectors to the molecules and transpose them to a 
single origin [see figure 25(a)], the number of velocity vectors ending in the 





arpa -- dn 

area — 






Figure 25. Velocity vector diagrams (a) for molecules with velocities between 
c and c+du+dv+dw, (b) velocity distribution for a one-dimensional gas. The 
area of the shaded band gives the fraction of molecules with velocities between 
u and u+du in the u direction. 

element du, dv, dw gives us the number of molecules with velocities between 
u and u+du,v and v + dv, w and w+dw. This number is then 

dn = A exp 

\ 2kT} 

du dv dw. 


Before proceeding to use this relation we may split up c 2 into its three 
components u 2 + v 2 + w 2 so that 

dn — A exp I | expj ) expl \ dudv dw. (5.41) 

\ 2kT) v \ 2kT) v \ 2kTf 

Since each of the three components must have the same velocity distributions, 
the quantity 

/4* exp 

/ mu 2 \ 
\ 2kT} 


gives the number of molecules with velocity between u and u + du irrespective 
of the v and w components. The fraction of molecules in this category is 

78 Gases, liquids and solids 

^ = Bexp (_^!\ dM . (5.42) 

The constant B may be determined by observing that, if this relation is 
integrated for all possible velocities from - oo to + oo, the resulting fraction 
must be unity. Then 

J>"(-3?)*- L *"» 

Putting -^- = a (5.44) 






dn I m \* / m« J \ , ._ ._ 

From the table of integrals at the end of the chapter 

= z i e — au = I - 1 = 

Jo W 

Hence B = (V^Y • 

How can we plot this function ? We bring the term du over to the left-hand side 
so that we have 

dn 1 

— —or 
n du 

1 dn ( m \* / mu 2 \ „ ,_, 

^ = M exp (-2*r)- (546) 

We may calculate the right-hand side and plot it as a function of « [see figure 
25(6)]. If we take a strip of width du its area is then dn\n which is the fraction 
of molecules with velocity between u and u+du. We see that the greatest 
fraction occurs for zero velocity; this is because of the exponential nature 
of the Boltzmann factor, and the fact that the 'one-dimensional' gas directly 
reflects this distribution. 
The average energy in the u direction is given by 

2 \2xkTJ J _«, 2 \jt/ J _„ 

79 Further theory of perfect gases 

From the table of integrals at the end of the chapter 

Hence . - = (•)*! (•)*„ 5*1 - 
2 \n] 2 \a 3 } 4 a 

4 m 


or i = y . (5.48) 

Similarly the fraction of molecules with velocity components between 
u and u+du and between p and v+dv is simply the product of the two 
individual probabilities. 

Finally the fraction of molecules with velocity components between u and 
u+du, v and v+dv, w and w+dw is 

Generally we are not interested in the dependence of « on the individual 
components of velocity. For practically all purposes we wish rather to know 
the number of molecules with velocities between c and c+dc, irrespective of 
direction. We may calculate this easily by drawing a sphere of radius c and 
another of radius c+dc [see figure 26(a)]. The number of velocity vectors 
ending in the spherical shell is then the number required. In velocity space the 
volume of the shell is 4nc 2 dc and this replaces du do dw in equation (5.50). 
We then obtain 

ldn I m \* 

-— = 4wc* \-—r~ I < 
n dc \27ikT) 

\ 2kTj' 

-— = 4*c* -^-) exp|-^-i. (531) 

This is plotted in figure 26(6). We note that in any integration to give total 
quantities, c ranges from zero to oo, not from — oo to + ao. 

[The reader, if he is interested only in the three-dimensional gas, may pass 
at once from equation (5.40) to equation (531) by writing equation (5.40) 
in the form 

— = C exp | -——I AmPdc. 
n \ IkT) 

J'™ 00 dn / m \i 

— = 1, hewillfind that C has the value |-^— I . 
e .o n \2nkT) 

This results in equation (531).] 

80 Gases, liquids and solids 

5-4-2 Average velocities 

Rewriting equation (5.51) in the form 

dn (a\* 



and using the table of integrals at the end of the chapter we may readily 
calculate the following three representative velocities: the most probable 
velocity c m , that is the velocity for which equation (5.52) is a maximum, the 
average velocity c", and the root mean velocity (c 2 )*. We have 

most probable 




Jo n V n\m) 





Before leaving this we note one of the confusing features of probability 
distributions. According to equation (559) the population density is a 
maximum when c = 0. This is also true, as we saw above, of the velocity 
distribution in a 'one-dimensional' gas. As soon, however, as we specify 
velocity elements in two or three dimensions the most probable velocity 
(as distinct from the probability density) no longer corresponds to zero 
velocity. This is specifically brought out in equation (551) and figure 26 for a 




velocity c 

Figure 26. (a) Velocity vector diagram for molecules with velocities between 
c and c+dc. This is given by the number of points ending the velocity vectors, 
which lie within the spherical shell of radii c and c+dc. (b) Maxwellian velocity 
distribution for a three-dimensional gas. The shaded band gives the fraction of 
molecules with velocities between c and c+ dc. The approximate positions of the 
most probable velocity cm. the mean velocity 6 and the root mean square velocity 
(?)* are indicated. 

81 Further theory of perfect gases 

three-dimensional gas. A similar point will arise when we discuss the behaviour 
of rubber molecules in Chapter 8. 

5-4'3 Number of molecules striking a surface per second 

The number of molecules striking a square centimetre of surface per second 
may be directly obtained from equation (5.45). If the surface considered is 
at right angles to the u direction the v and w components are not involved. 
Consider molecules with velocity u, u+du; we construct a prism of unit cross- 
section and length u. Then all the molecules in this volume will reach the 
surface in one second. This number is u dn where dn is the number per c.c. 
with velocity « in the sense towards the surface. The total number required is 
the integral of u dn for all values of u from to infinity: this choice of limits 
ensures that we are considering only those molecules moving towards the 

The number is 

/ kr\j = « /8 x m* 

\2nmJ 4\nmJ 
Comparison with equation (5.53b) shows that this is equal to inc. 

5.4.4 Maxwell's derivation of the velocity distribution in a gas 

The velocity distribution derived above makes use of the Boltzmann dis- 
tribution. It is interesting to see how Maxwell derived the velocity distribution 
before the more general ideas of Boltzmann had been developed. The 
following is based on Maxwell's paper published in 1860. 

Let n be the total number of molecules, and let u, v, w, be the components 
of the velocity of each molecule in three rectangular directions. Then if the 
number of molecules for which velocity component u lies between u and 
u+du is nf(u) du, where /"(a) is the function to be determined, the number of 
molecules for which v lies between v and v + dv will be nf(v) do and similarly 
for w, where /always stands for the same function since there is no preferred 
direction in the gas. 

Now the existence of the velocity n does not in any way affect that of the 
velocities v and w since these are all at right angles to each other and 
independent. Consequently the number of molecules whose velocity com- 
ponents lie simultaneously between u and u+du, between v and v+dv and 
between w and w+dw is 

82 Gases, liquids and solids 

dn = nfW/Wfiw) dudvdw. (S- 548 ) 

Let this now refer to all those molecules which have a resultant velocity 
c where 

c 1 = u 2 +v 2 + w 2 . ( 554b > 

We now consider the condition under which the components u, v, w, can vary 
whilst c remains constant. This is found by differentiating equation (5.54b) 
and putting dc = 0; we obtain 

u du+v dv+w dw = 0. (5.54c) 

Since no direction is preferred over any other, it follows that dn in equation 
(5.54a) must remain constant whatever the individual values of u, v and w, 
provided these satisfy equation (5.54c). This implfcs that 

— (dn)du+— (dn) dv+4- (dn) dw = 0. (5.54d) 

du dv dw 

From equation (5.54a) this gives 

/'(«) duf(v)f(.w)+f'(v) dof(u)f(w)+f'(w) dwf(u)f(v) = 

where/ '(«) is the differential of/(«) with respect to u, etc. Dividing throughout 
by/"(«)/(")/O*0 we obtain 

ZW^i+ZW*^ (5.54e) 

mm m 

In order to solve equations (5.54c) and (5.54e), we multiply equation (5.54c) 
by an arbitrary constant X and add it to equation (5.54e). The result is 

L/(«) J If & i L/w J 

Each term must identically be equal to zero, and since du, dv, dw, although 
very small are not themselves zero, the quantities in the brackets must be zero. 

IM +Xu = o or ?&=-*. 
/(«) /(«) 

Integrating this relation we have 

where A is an integration constant. This may be written in the form 

83 Further theory of perfect gases 

As Maxwell remarks, If X were negative, the number of molecules would be 
infinite, so consequently, X must be positive. The number of molecules with 
velocities between u and u+du then becomes 

dn = nf(u) du = nB exp 


There is no way of establishing the value of X except by calculating the 
mean square velocity c 5 and making use of the Gas equation: \nv? = kT. 
It is then found that X has the value m/kT. The result is 

dn = «B*Kp\-—)d«, 
which is identical with equation (5.42). 

5 -4-5 Experimented determination of velocity distribution 

In most of the earlier experimental studies of the velocity distribution, a 
furnace was used to produce a vapour of metal atoms, which behave like a 
monatomic gas, at the temperature of the enclosure. They were allowed to 
condense on a cold surface and by the ingenious use of moving shutters atoms 
with different velocities were caught at different points on the surface. The 
intensity of condensed atoms provides then a measure of the relative number 

velocity c 

xxx experimental 
— — theoretical 

Figure 27. (a) Method due to Zartman and Ko for determining the velocity dis- 
tribution of bismuth atoms; the atoms are evaporated, collimated by the slits 
Si, S 2 , S 3 and enter the rotating drum. They fall on the cold plate P where they 
condense. (6) Results obtained in a later experiment by Esterman, Simpson and 

84 Gases, liquids and solids 

of atoms within that velocity range. One method due to I. F. Zartman (1931) 
and C. C. Ko (1934) is illustrated in figure 27(a). Atoms of bismuth are pro- 
duced by a furnace and the vapour is collimated by a series of slits, Si, S 2 , S 3 ; 
the vapour beam reaches a drum rotating at a speed of 6000 rev. min. -1 and 
can only enter the drum at the slit S. The atoms then strike the plate P where 
they condense; the fastest ones reaching A, and slower ones B and the slowest 

A more elegant and refined experiment by I. Esterman, O. C. Simpson and 
O. Stern (1947) gave results shown in figure 27(6). The agreement with the 
Boltzmann distribution is surprisingly good. 

5-S Thermal energy of molecules 

5*5*1 Specific heats, number of degrees of freedom 

We have already seen that for a perfect gas in thermal equilibrium each 
molecule possesses an average thermal energy of translation of amount 
§ kT. Since a molecule has 3 independent degrees of translational freedom 
we may associate energy \kT with each degree of freedom. This may be gen- 
eralized to apply to all particles in thermal equilibrium; however, for macro- 
scopic bodies the resulting velocity is rather small. For example the mean 
thermal velocity of a cricket ball in equilibrium with its surroundings at 
room temperature (say 300°K.) is about 10 -8 cm. sec. -1 . 

The result may be generalized even further. If we define the number of 
degrees of freedom as the number of independent squared terms that enter 
into the total energy of a particle, then each has associated with it an average 
thermal energy of amount $kT. For example the kinetic energy due to trans- 
lation is determined solely by (linear velocity) 3 in each of these independent 
co-ordinates; each degree of freedom has thermal energy \kT. The energy of 
rotation of a body about a specified axis is determined by (angular velocity) 2 , 
so that each of these degrees of freedom has energy \kT. Vibrations present 
a special problem. If the equation of motion, for a body of mass m, is 

x = -aPx, with solution x = Asincot, (5.55) 

the kinetic energy at any instant is -Jm* 2 , whilst the potential energy is easily 
shown to be equal to ^maPx 2 . By substituting in equation (535) we find that 
the sum of kinetic and potential energy is \mA 2 a> 2 , so it is independent of x 
and x. Nevertheless at any instant the kinetic energy is proportional to 
x 1 and the potential energy to x 2 . Each of these contributes $kT to the average 
thermal energy. 

Summarizing we may write: thermal energy per degree of translational 
freedom = \kT, per degree of rotational freedom = ^kT and per vibrational 
mode (equivalent to two degrees of freedom) = kT. 

86 Further theory of perfect geses 

5-5-2 Internal energy of a gas 

If we consider the energy which arises solely from thermal motion (not 
excited electronic states), the internal energy u per molecule of a gas may 
be deduced for monatomic, diatomic and polyatomic molecules. 

Monatomic gas: only 3 translations u = f fcTper molecule. 

u = %kT per molecule plus any 

Diatomic gas: 3 translations) 
2 rotations I 

vibrational energy. 

The third rotation about the axis of the molecule does not occur for reasons 
to be described later. 

Polyatomic gas: 3 translations I 
3 rotations J 

These results are summarized in Table 14, 

u = f IcT plus any vibrational energy. 

Table 14 Internal energy and specific heats of gases 



















«mol -1 




E/g-mole -1 








C, = C r +R 





7 C r 






C r cal. g-mole" 








H 2 : 4-9 

H 2 S :6 

He: 3 

2 : 50 

N 2 : 4-6 
Cl 2 :6 

CS 2 :10 

For monatomic gases the agreement is excellent. For diatomic gases the 
agreement is good, but the value given for chlorine seems to indicate that 

86 Gases, liquids and solids 

it possesses additional vibrational energy at room temperature. With poly- 
atomic molecules the behaviour is very complicated but it would seem that, 
in general, CV > 6 calorie per gram-molecule. 

In general if one knows the number of degrees of freedom / which are 
'active* one can write 

so that C v - \Rf 

and y = Q=l+l. (5.56) 

C r f 

5-5-3 Gas kinetic theory ofadiabatic expansion 

Before we go on to discuss in greater detail the vibrational and rotational 
energy of a molecule, we may at once use the ideas expressed in the previous 
section to deduce the adiabatic gas equation for monatomic and polyatomic 

In an adiabatic expansion all the energy is taken from the internal energy 
of the gas. We first consider a monatomic gas where the internal energy arises 
solely from the kinetic energy of translation. 

From the kinetic theory we have: 

PV = $ (total kinetic energy Z) 

Z = \PV = internal energy U. 

The work done by the gas in an adiabatic expansion is 

PdV= -dU= ~dZ = -dQPV). 

.-. pdV= -\PdV - \VdP 

or fPdV+j;VdP = 

c dV ^dP „ 
or 5 — +3 — = 0. 

V P 

Hence PV* = constant. (5.57) 

For a diatomic gas possessing two rotational degrees of freedom, the internal 
energy is 

V = kinetic energy + rotational energy 

= \PV+RT=\PV+PV 

= |PK. 

87 Further theory of perfect gases 

£ = /i(7H-l) — , (5.59) 

For an adiabatic expansion PdV = -d(%PV). This leads, as above, to 

PV 115 = constant. (5.58) 

5-5'4 Rotational and vibrational degrees of freedom 

At room temperature the specific heat Cy of hydrogen is about 5 cal. mole' 1 . 
As the temperature is lowered C r decreases until at - 220°C, C r has a value 
of only about 3 cal. mole -1 . It is apparent that the rotational degrees of 
freedom have gradually disappeared and that at about 50°K. the molecule 
has only translational energy. The reason for this is that according to the 
quantum theory rotors can possess only discrete energy levels. For a sym- 
metrical diatomic molecule rotating about an axis, for which the moment of 
inertia is /, the possible energy levels are 

h 2 

where h is Planck's constant (6-6 x 10 -17 erg sec. -1 ) and n is an integer. The 
smallest value of rotational energy permitted occurs when n = 1 and has 
the value 

(£)»i.= -^-. (5.(50) 

For a hydrogen molecule, /about an axis normal to the bond is approximately 
4-7xl0- 41 

.'. (£r)m t » = 2x 10- »« erg. 

But the translational energy E, = $kT = 2 x 10 _16 rerg, so that at 50°K. 

£;=10- X4 erg. (5.61) 

Therefore the translational energy is appreciably less than the lowest rota- 
tional energy level. By Boltzmann's principle the fraction of molecules 
capable of acquiring rotational energy 2x 10~ 14 erg at 50°K. will be 

Consequently below 50 C K. very few of the molecules will take up rotational 

motion. At room temperature (300°K.), kT will be large compared with the 

first rotational energy level, so a large fraction of the molecules will be able 

to take up the first and higher rotational energy levels. 

Equation (5.60) also shows why rotating about the axis itself does not 

occur. The moment of inertia / is so small that the lowest rotational energy 

h 2 
state, — — , is enormous and can never be reached before thermal dissociation 


88 Gases, liquids and solids 

Similar considerations apply to the vibrational degrees of freedom. If the 
vibrational frequency is v, the first energy state (ignoring zero-point energy) 
demands energy of amount equal to hv. For the hydrogen molecule 
v a 2-6 x 10 14 sec. -1 , so that hv is about 2x 10" " erg. For an appreciable 
fraction of molecules to be excited to this level fcrmust have about the same 
value, so that Tmust exceed 10,000°C. The molecule dissociates long before 
this. By contrast the chlorine molecule which has a much lower natural 
frequency v, is able to take up appreciable vibrational energy at temperatures 
above 600°C. so that its specific heat approaches C v = 7 cal. g-mol." 1 at 
elevated temperatures (see figure 28). 





liquifies ^^^ 




/ normal H 2 











i i ■ i i 

■ j i i 

10 20 50 

temperature "K. 

1 00 200 

500 1000 2000 

5000 10000 

Figure 28. Variation of specific heat with temperature for H a and for Cl». 

S -5 -5 Calculation of vibrational energy as a function of temperature 

We reproduce here the standard treatment for a simple oscillator of natural 
frequency v. According to quantum theory the possible energy levels are 

E. = («+i)Av, 


where « is an integer. 

The zero-point energy -JAv is the vibrational energy the molecule possesses 
even at absolute zero, but since this is a constant we may ignore it at this 
stage of the computation and add it at the end. We put 



89 Further theory of perfect gases 

Applying the Boltzmann function, the number of states with quantum number 
n is 

XT 1 "fa 


= A exp (-n0) where - — . (5.65) 

The total number N of atoms is the sum of all JV„s from n = to n = oo , 
N = A[\ + exp (-0) + exp (-20) + exp (-30)+ . . .] 
= A[l + exp(-jg) + (exp(-/0) 2 + (exp (-£))* + . . .] 

= - . (5.66) 


This gives .4 so that JV. becomes 

N„ = N[l - exp (-j8)] [exp (-«/?)]. (5.67) 

The total energy is the sum of NJ5 y where E v = nAv. 

U => 7V[1 - exp (-0)] [exp (-«/?)] «Av summed for all values of « 
= Ml - exp (-0)] {hv [exp (-0)] + 2hv [exp (-£)] 2 + 
+ 3Mexp(-0)] 3 +...} 

= N[l - exp (-0)] hv [exp (-)?)] {1 + 2[exp (-0)]+ 
+ 3[exp(-j8)] 2 +...} 


7VT1- exp (-0)] hv [exp (-£)] 

[1 -exp (-0)1 


= M, fc*P<-^ = _^_ . (5.68) 

[1- exp (-/?)] exp (/?)-! 

If we include the zero-point energy %hv we have 

r/ = Nhv{i+ [exp C8)- 1]- 1 }. (5.69) 

.... -. hv . hv . 

At high temperatures exp = exp — ~ 1 +— so that 

U aNhv 

ri Arri 

1/ = iMv+NkT. (5.70) 

Thus apart from zero-point energy each vibration carries average energy 
kT, so that the specific heat contribution is Nk = R for each vibration. At 


lower temperatures exp J — I — 1 1 and therefore the specific heat falls off 
90 Gases, liquids and solids 

and tends to zero when T = (see figure 29). This decrease in vibrational 
energy with temperature reproduces essentially the right-hand features of the 
specific heat curves of hydrogen and chlorine in figure 28. We may also note 






Figure 29. Variation of vibrational specific heat with temperature for a simple 
oscillation of natural frequency v. Specific heat is in units of R per g-mole; 

• . k 
temperature in units of j-. 

in passing that this concept was applied by A. Einstein to explain the specific 
heat of solids. He suggested that each atom has 3 independent vibrational 
modes of constant frequency v. The specific heat (in this case it assumes no 
volume change and is therefore C F ) is 3R at high temperatures and falls to 
zero at low temperatures. We shall discuss this in greater detail in a later 

5 • 5 -6 Rotational energy 

For a symmetrical diatomic molecule rotating about an axis for which the 
moment of inertia is J, the possible energy levels E T are as given in equation 
(5.53). The application of the Boltzmann distribution is complicated because 
each state of quantum number n consists of g states of practically identical 
energy. The degeneracy factor g has the value 

g = 2«+l. 


One can build up the Boltzmann relation as before where the basic exponential 
term is exp J J or exp I -n(/i+ 1) — — —J . The final integral has to be 

91 Further theory of perfect gases 

summed numerically. For high temperatures the energy for N rotors becomes 
NkT. This is the classical result for a rotor with two degrees of freedom. In 
general the rotational energy falls off from Aifcrwhen 

kT< £i- **» 

5*5*7 Translationa! energy 

For a gas confined within a container, quantization of translation^ energy 
does occur, but the number of energy-states is enormous and the spacing 
between energy levels is minute. We may therefore consider all energy states 
to be available as a continuous distribution, so that the classical Boltzmann 
distribution applies. Consequently each degree of translation^ freedom 
involves energy ikT per molecule under all practical conditions. 

5 -6 Macroscopic examples of equipartition of energy 

We conclude this chapter with four examples of thermal motion occurring 
in systems other than gases. 

5-6-1 Galvanometer mirror 

A delicately suspended galvanometer mirror can be seen to undergo small 
random oscillations which are due to its thermal energy. The mirror has a 
single degree of oscillation about its axis so that we attribute to its mean 
square velocity (cP), energy ikT, and to its mean square angular deflection 
(W 1 ), energy ikT. We may write 

ikT = i/(S5) = ir (P), (5.73) 

where / is the amount of inertia, t is the torsion constant. If we determine the 
torsional stiffness of the suspension we may then calculate (?»). This tells us 
the amount by which any galvanometer reading will fluctuate as a result of 
thermal motion. These fluctuations do not depend on the presence of air 
surrounding the mirrors; it is true that gas molecules will buffet the mirror 
and contribute to the oscillation, but they will also contribute to the damping, 
so the net effect is unchanged. The mirror behaves like a large molecule with 
a single mode of torsional oscillation. In a perfect vacuum the oscillations 
can be considered as arising from the random absorption and radiation of 
electromagnetic energy associated with the temperature T. The resulting 
behaviour is the same. 

3-6-2 Sedimentation 

As we saw earlier particles in thermal equilibrium behave like gas molecules 
of large molecular weight. If they are in an evacuated container the sedimen- 

92 Gases, liquids and solids 

tation due to gravity is so rapid that they virtually all lie at the bottom of the 
container. If, however, they are suspended in a liquid of very nearly the same 
density the effective gravitational field is greatly reduced and there may be an 
appreciable variation in particle concentration with distance from the bottom 
of the container as given by equation (535). 

5-6-3 Electrical noise 

The conduction electrons in a metal may be regarded as a gas with random 
velocities. These give rise to electrical 'noise'. The root mean square potential 
fluctuation across a resistor of resistance SI is given by 

r.m.s. potential = </[{4QkT(f t - A)], (5.74) 

where f 2 -/I is the bandwidth frequency over which the measurements are 
made. This result at once shows that there is a limit to useful amplification. 
If the original signal is too small (e.g. comparable with the random electrical 
noise) no amount of amplification will improve reception. 

Similarly in the thermionic emission of current /<> it may be shown that there 
is a fluctuation / in the emitted current given by 

|P|*= lfe\, (5.75) 


where e is the electronic charge and t the time constant of the measuring 
system. This effect has indeed been used as an independent means of deter- 
mining e; the result agrees with other methods to within 1 per cent. 

5-6-4 Brownian motion 

The English botanist R. Brown observed in 1827 that finely divided particles 
in suspension undergo ceaseless movement. This motion was analysed by 
Einstein in 1906; the following derivation is based on a treatment developed 
by P. Langevin in 1908. We first assume that the particles possess thermal 
kinetic energy in one dimension of amount \kT. 

If at any instance the molecules of the liquid collide with an individual 
particle to produce a net force X in the x direction, we may write the equation 
of motion of the particle in the form 

mX+6xatix = X, (5.76) 

where m is the mass, a the radius of the particle (assumed spherical) and n the 
viscosity of the fluid. In this equation the first term represents the inertia! 
force on the particle and the second the frictional force due to the viscous 
resistance of the fluid; this is Stokes' well-known relation for viscous forces 
on a sphere. The third term represents the force due to molecular collisions. 
Thus from the point of view of collisions we treat the fluid as a collection of 
individual molecules, but from the point of view of the resistive forces we 

93 Further theory of perfect gases 

treat the fluid as a continuum. Finally this equation refers only to motions of 
the particle produced by collisions; the intrinsic motion of the particle by 
virtue of its own kinetic energy i^kT) emerges from the analysis. 

We multiply both sides of equation (5.76) by x and then average over a 
large number of collisions. Then for any fixed specified value of x, for every 
value of X there is an equal chance of — X occurring so that "LxX = 0. 

I.mxx+6narjl,xx = XxX = (5.77) 

or m xx+6natj xx = 0. 

We note that — (xx) = xx+x 1 = xx+u* where « is the velocity in the x 

direction. Then 

m — xx— mu 2 + (mat) xx = 0. (5.78) 


It turns out that the first term implies a transient at the beginning of the 
collision which has a negligible effect on the subsequent displacement.* 


Gnat] x~x = mi? = kT, (5.79) 

or 3nati - (F) = kT. (5.80) 


In a time interval x the mean square displacement Ax* is then 

._ kTz RTt 

A* 2 = -z = z — . (5-81) 

iJiaij inatjNo 

where N is Avogadro's number. A given particle is observed say every 
30 seconds and the displacement Ax in the x direction is observed for each 
interval of time. These values are squared and the mean value is calculated; 
this gives Ax*. This is an average path for the time interval x = 30 seconds. 
The value of Aj? so found may be compared with the right-hand term of 

* More precisely we may write xx as j> j:(* 2 ) which we may call z. Equation (5.78) 
then becomes 

mi + 6nanz = mu 1 = kT. 
The solution is 

kT , ( -6nar(t\ 

where A is a constant of integration. The quantity — — is extremely large so that even 
for minute values of t the exponential term becomes negligible. During the total time of 
the collision, z is thus given essentially by the first term on the right-hand side of the equation. 
This is indeed the same equation as (5.80). 

94 Gases, liquids and solids 

equation (5.81). Experiments show that the results agree very well with 

There are two points of interest. First the displacement does not depend 
on the mass of the particle (only on its radius). Secondly although the 
individual particle has intrinsic motion, which has nothing to do with the 
surrounding liquid, the distance it moves about an equilibrium position 
is determined by the viscous damping of the liquid. In a descriptive way we 
may say that a viscous force (marjji acting over a distance x consumes the 
thermal energy imu 1 . The particle then reacquires its thermal energy and the 
process is repeated. Apart from a numerical factor this is essentially the result 
implied by equation (5.79). 

If then equation (5.79) is a description of the way in which viscous damping 
consumes thermal energy, is it not possible to apply it to the behaviour of a 
gas molecule within its own gas? We have already derived a relation for the 
viscosity of a gas, namely t\ = \mncX. However this can only be used when 
the distances between the moving bodies is larger than the mean free path L 
This implies that Stokes' law is valid for gases only if the radius of the sphere 
is larger than A. Thus we cannot really use Stokes' law if the sphere is the gas 
molecule itself. However as an extreme case let us see what happens if we 
take a heavily condensed gas where A = a. We have 

q = \mnca; x ^ u; x ^ a. 
Then the L.H.S. of equation (5.79) becomes 

(matixx = dna\mnca.u.a = 2na 3 nmcu. 

But c ~ V3(«). 

Hence 6nanxx <z \0a 3 nmu*. (5.82) 

But each sphere of radius a occupies a cube of volume 8a 3 , so that 8a 3 » = 1 . 
The value of equation (5.82) is thus approximately mu 2 , which agrees with 
the r.h.s. of equation (5.79). This is, of course, pushing the model to an 
extreme, but it shows again that the particle considered in the analysis of 
Brownian motion is essentially no different from a molecule, except that it is 
very much larger. 

This has been a long chapter, and we now conclude it with a relevant 
quotation from Lucretius: 

This process, as I might point out, is illustrated by an image of it that is continually 
taking place before our very eyes. Observe what happens when sunbeams are 
admitted into a building and shed light on its shadowy places. You will see a 
multitude of tiny particles mingling in a multitude of ways in the empty space within 
the light of the beam, as though contending in everlasting conflict, rushing into 
battle rank upon rank with never a moment's pause in a rapid sequence of unions 
and dis-unions. From this you may picture what it is for the atoms to be perpetually 
tossed about in the illimitable void. To some extent a small thing may afford an 
illustration and an imperfect image of great things. Besides, there is a further reason 

98 Further theory of perfect gases 

why you should give your mind to these particles that are seen dancing in a sunbeam : 
their dancing is an actual imitation of underlying movements of matter that are 
hidden from our sight. There you will see many particles under the impact of invisible 
blows changing their course and driven back upon their tracks, this way and that, 
in all directions. You must understand that they all derive this restlessness from the 
atom. It originates with the atoms, which move themselves. Then those small 
compound bodies that are least removed from the impetus of the atoms are set in 
motion by the impact of their invisible blows and in turn cannon against slightly 
larger bodies. So the movement mounts up from the atoms and gradually emerges 
to the level of our senses, so that those bodies are in motion that we see in sunbeams, 
moved by blows that remain invisible. 

(It is interesting to note that, almost two thousand years later Perrin used 
the identical argument in his studies of Brownian motion of gamboge 
particles in a liquid: he suggested that this could be projected optically before 
an audience to demonstrate the molecular movement within the liquid.) 

Prof. S. Sambursky in his perceptive study The Physical World of the 
Greeks adds the comment that this remarkable description, 'perfectly 
describes and explains the Brownian movement by a wrong example. The 
movements of dust particles as seen by the naked eye in sunlight are caused 
by air-currents : the real phenomenon postulated by Lucretius on the basis of 
abstract reasoning can only be seen in a microscope.' Further, as we have 
pointed out previously, the random motion of fine particles is an intrinsic 
property of any assembly of free entities in thermal equilibrium, whether they 
be gas molecules or solid particles. Nothing, however, can detract from the 
beauty and elegance of this inferential type of argument. 


Values of \ x*e-"'dx 

xe-"\fc = — 
Jo 2a 


x 3 e-"'dx = -i- 
o 2^ 

For all these integrals 

J -oo Jo 

96 Gases, liquids and solids 

Chapter 6 
Imperfect gases 

Many gases even at room temperature deviate markedly from the 'ideal* gas 
relation PV - RT. Similar deviations are observed with noble gases at higher 
pressures and lower temperatures. In this chapter we shall discuss the 
behaviour of such *imperfect' gases and show that their properties can be 
explained in terms of the 'ideal' gas theory, modified to allow for the finite 
volume of the gas molecules and for the existence of intermolecular forces. 

6-1 Deviations from perfect gas behaviour 

6*1-1 The virial equation 

In general, instead of the relation PV = constant at a fixed temperature, 
imperfect gases obey an empirical power equation of the form 

PV=A + BP+CP'+ DP 3 . (6.1) 

This is known as a virial equation and A, B, C, and D as virial coefficients. 
The most important coefficient, apart from A, is B, since C and D are usually 
very small. At low temperatures B is negative, at higher temperatures positive 
and at some intermediate temperature T B it has the value zero. At this 
temperature, if the pressures are not too high we see at once that equation 
(6.1) reduces to 

PV = A = constant, (63) 

that is Boyle's law becomes approximately true again. For this reason T B is 
called the Boyle temperature. 

Plots of PV against P are shown schematically in figure 30. If we measure 
the slope of the curves near thePFaxis, i.e. near P = we have 

«rn =B+ 2CP + ... 


= Bfor P — 0. (6.3) 

The slope near P = will, therefore, be zero at the Boyle temperature. It is 
also easy to show that the curve connecting the minima is a parabola. 

97 Imperfect gases 


Figure 30. Variation of PV with P for a real gas. At one special temperature where 
the initial tangent is horizontal PV is constant at the lower pressure range ; this is 
the Boyle temperature 7" . 

6-1-2 Andrews' experiments on carbon dioxide 

T. Andrews (1813-1885) carried out a series of important and classical 
experiments on the compressibility of gases and the conditions under which 
they can be liquefied. His first paper appeared in 1861 and one of his major 
surveys was given in his Bakerian lecture of 1869. We summarize his main 
conclusions as exemplified by the behaviour of C0 2 (see figure 31). 

Figure 31. Variation of pressure P with volume V for C0 2 (based on Andrews' 

98 Gases, liquids and solids 

(i) Above a temperature of about 48°C, the compressibility of C0 2 

resembles that of a perfect gas. 
(ii) At, say, 21 °C. compression produces liquefaction. 
From B to C the material behaves as a gas, from C to D it condenses at a 
constant pressure, and at D it is completely liquid. From D to E the slope is 
very steep because a liquid is relatively incompressible, 
(iii) At 31 1°C. there is no liquefaction although the isotherm is very 

distorted from the ideal gas isotherm, 
(iv) At 30-9°C. liquefaction just occurs under compression. 
Above this temperature, liquefaction cannot be produced. This is known as 
the critical temperature T c , the temperature above which liquefaction cannot 
be achieved however high the pressure. The pressure at which liquefaction 
just occurs is called the critical pressure P„ and the corresponding volume 
(for one gram-molecule of the substance) is called the critical volume V c . 

6-1 -3 Continuity of liquid and gaseous state 

Let us consider two isothermals, one above T c and one below. Consider the 
points J and K on these isothermals at some pressure level above P c . At J 
the substance is purely a gas, at K purely a liquid; this means that if we keep 
the pressure constant at J but steadily reduce the temperature we may pass 
to the liquid phase at K. This, in turn, implies that above P c it is possible, 
simply by cooling, to pass from the gaseous to the liquid state without any 
mixture of phases. 

6-1-4 Significance of T c 

The critical temperature may be easily understood in terms of intermolecular 
forces. Suppose figure 32 represents the potential energy between one molecule 





"of separation 

Figure 32. Potential energy between one molecule and a 
single neighbour as a function of separation. 

99 Imperfect gases 

and a single neighbour, as a function of separation. It is clear that if the 
thermal energy of the molecule is greater than Ac, one molecule can always 
escape from its neighbours. However closely they are pressed together 
(within limits) the thermal energy will always be sufficient to overcome the 
attraction between the molecules. 

Now the thermal energy of each molecule is of order kT where k is 
Boltzmann's constant and equals l-4x 10 _1<s erg. deg. -1 and 7* is the absolute 
temperature. Thus a molecule can always escape from its neighbour if 
kT> Ae 

or T >J- (6.4) 

The substance is therefore unliquefiable for temperatures above 

Some typical results are given in Table IS. It is seen that the agreement is 
good, indeed surprisingly good, in view of the extremely simple assumptions 
made. This model deals only with the problem of a molecule escaping from a 

Table 15 Critical temperatures 

Gas As ergs — °K T t (observed) 

(theoretical) k °K. 


0-8 x 10" 15 



H 2 




N 2 




single neighbour. In section 6-2-6-2 we shall consider this behaviour in 
somewhat greater detail and, using a different model, establish essentially 
the same results. 

6-1 -5 The miscibUity of liquids and gases 

The continuity of the liquid and gaseous states has an interesting bearing on 
the solubility or miscibility of liquids. Our starting point is that, unlike solids 
and liquids, all gases are completely soluble in one another (see p. 46). Let 
us now consider the behaviour of, say, water and oil: these are mutually 
insoluble in the liquid state and if mixed together will separate into two 
distinct phases. However, water vapour and hydrocarbon vapour form a 
homogeneous mixture in the gaseous phase, in which there is a perfect random 

1 00 Gases, liquids and solids 

mixing of molecules of the two materials. If now the vapour is very highly 
compressed it can be brought to a density of the ordinary liquid ; but provided 
the temperature is high enough -above the 'effective critical temperature' of 
the system - it will behave as a gas and still remain as a single-phase material. 
We shall thus have achieved complete miscibility of oil in water. Of course, 
this can be seen from a completely different angle. If water and oil are enclosed 
in a container at constant volume and heated, a temperature will be reached 
at which the thermal energy is sufficient to overcome the intermolecular 
attraction of like and unlike molecules - the material will then cease to consist 
of two separate phases and complete miscibility will have been achieved 
Alternatively we may say that by increasing the temperature the entropy of 
the system is increased and the randomization which this implies may be 
sufficient to mix the materials on a molecular scale, i.e. to achieve miscibility. 
Is the single-phase material now a high-density gas or a high-temperature 
liquid? To some extent this is a matter of semantics. 

6-2 Kinetic theory of an imperfect gas: van der Waals equation 

6-2-1 Derivation of van der Waals equation 

We shall now derive an equation for an imperfect gas, using the assumption 
that the main deviations from ideal behaviour arise from the finite size of the 
molecules and the forces between them. 

We first assume that if real and ideal gases are in thermal equilibrium, that 
is at the same temperature, the mean translational energy of the individual 

_ 3J?7* 

gas molecules (iiwc*) still implies a value of c 2 given by . This assump- 

tion, in effect, is a broader generalization of the discussion given on page 45. 

(Note that the Boltzmann distribution law, n t = Nexp I — - J, applies 

only if a includes both kinetic and intermolecular potential energy.) 

Size of molecule. We first deal with the finite size of the molecules. We may 
say that the collisions with the wall of the container which determine the 
observed gas pressure are exactly the same as for zero-size molecules, i.e . 
an ideal gas, but that the available volume is reduced by some amount b 
depending on the number and size of the molecules. We may, therefore, write 
for one gram-molecule 

P(V-b) m RT. (6.5) 

There are several ways of estimating b. 

(0 Molecular collisions between the walls. If a is the effective diameter of a 

molecule I volume v m = — J the mean free path between collisions is 

101 Imperfect gases 

reduced from k to k—a. Thus as the molecules travel between the walls of the 
vessel they will travel (A -a) between collisions where, on the ideal gas 
model, they would have to travel a distance L Thus the number of collisions 

on the wall per second is increased in the ratio . This means that the 


pressure is increased by the factor 7-7. Instead of the ideal gas equation 

PV = %Nm& = RT, 

we therefore have 

PV = ■ (6.6) 



P ( V ~ V l\ = * T ' (6-7) 

The correction term b in equation (6.5) therefore has the value 

b=V-= Va(na 2 n) 

= VflTtff 3 

= N 6o m (6.8) 

where we have used the simple value of k = — — derived in Chapter 4. On 

7KT 2 n 

this model, therefore, b is about 6 times the total volume occupied by the 

(ii) Collisions with the wall (Jeans). Jeans gives a rather tricky way of 
calculating b. He suggests that all molecules are surrounded by a sheath of 
radius a which excludes the other molecules. The excluded volume is therefore 
8 times the volume of the molecules themselves. For a gram-molecule con- 
taining No molecules this is therefore 8 N v„. There is thus a zero chance of 
finding a molecule within the excluded volume and a chance 


V-W v, 

of finding it in an element of volume d V outside the excluded volume. He 
then considers the space available for molecules about to collide with the 
wall and argues that, since only the farther halves of the spherical sheaths are 

102 Gases, liquids and solids 

excluded when collisions are imminent, the chance of dV being free in space 

— . (6.9b) 

The chance of a molecular collision with the wall is found by multiplying 
the two probabilities together. The result is 

dV V-4N v„ dV V 

V V-8N»v m V V-4N v m 

For a gas of negligible size the probability is — so that for the real gas the 

collision probability, and hence the pressure, are increased by the ratio 


. This is equivalent to replacing Kby V-4N v m in the ideal gas 

V-4N v m 

equation so that 

b = 4N v m . 

Readers who find it difficult to appreciate the subleties of this treatment may 
prefer the following. 

(iii) Excluded volume. If a molecule approaches another within a distance a 
between centres contact occurs. Thus each molecule appears to carry a sheath 
of radius a, i.e. of volume 8v m which excludes other molecules. If, therefore, 
we start with a container of volume V, the first molecule has a free volume V; 
the second a free volumesof V— 8v m ; the third a free volume of V— 2 x 8^; 
the fourth a free volume oPyp'— 3 x Sv m . We then have to find an effective 
average value for all the N molecules in the gas. The standard procedure is 
to take the geometric mean, i.e. the Mh root of the product 

V{V- Sv m XV- 2 x 8v m W- 3 x 8v m ) . . . 
One thus obtains as the average available volume a value very close to 

so that b = 4N v m . (6.10) 

(iv) Excluded volume simplified. A much simpler approach (no less valid) 
is to say that each molecule carries an excluding volume of 8v m . But if we are 
concerned only with double collisions (ignoring triple and higher collisions) 
this means that on the average 8o m is the excluded volume for a pair of 
molecules. The average is 4v m so that once again we have 

b = 4N v m . (6.11) 

We have given these four different approaches to show that there is a 

103 Imperfect gases 

simple numerical factor relating the excluded volume to the volume of the 
molecule but that its value depends on the model used to derive it. Experi- 
ments agree well with a value of about 4 but readers should realize that this 
is a 'good' value, not a 'correct' value. 

6-2-2 Intermolecular forces 

We now turn to the effect of intermolecular forces. Within the bulk of the gas 
the molecular forces, on an average, act symmetrically on one another so 
that the net effect is zero. Consequently within the bulk of the gas the mole- 
cules behave as though they were in a gas without attraction, so that the 
effective pressure is the same as for an ideal gas P t . Near the walls of the 
vessel, however, the molecules have to escape from their neighbours before 
they collide with the wall. In overcoming this molecular attraction some 
kinetic energy is lost, the molecular velocity is reduced and the momentum 
imparted to the wall on impact and rebound is less than it would be for an 
ideal gas. Suppose the pressure defect is AP. Then the observed pressure P 
at the walls is less than P, by an amount AP. 
If we write the ideal gas equation, modified to allow for this, we have 

Ateal ^ideal = RT t 

or (P+APXK-6) = RT. (6.12) 

The pressure defect is proportional to the number of molecules striking the 
surface per unit area per second, i.e. to the density p. It is also proportional 
to the number of molecules attracting the molecules and reducing their 
impact. This is also proportional to p. As a result AP is proportional to p 2 . 
It could be written as aP 2 since P is nearly proportional to p, but it turns 

out that it is better to write this as — ; this correction term appearing to 
operate over a larger range than a term aP 2 . We may thus regard the cor- 
rection term — as the internal pressure in the gas, i.e. the pressure that has to 

be exerted to pull the molecules away from one another in overcoming 
intermolecular forces. Our final relation is 

(p+y\{V-b) = RT. (6.13) 

This celebrated relation is known as the van der Waals equation (published in 
1 873) and is a very convenient way of describing the behaviour of a 'real' gas. 
The model on which it is derived is admittedly a very crude one but it is 
generally felt that the equation holds better than it ought. Its main merit is 
that it is simple and that it allows for molecular volume and molecular 
attraction with the use of only two parameters a and b. 
If a and b are determined for a gram-molecule of gas, the van der Waals 

104 Gases, liquids and solids 

equation of state for n gram-molecules becomes 

6-2-3 Other equations of state 

There are a number of other equations of state which at various times have 
been proposed. R. Clausius (1822-1888) attempted to allow for the effect of 
temperature on the cohesive forces and suggested in 1880, 

but this involves three arbitrary constants. D. Berthelot in 1878 found that if 
he used a different value for a he could get about as good a fit with experiment 
by putting c = 0. He therefore used 

The best known 'rival' to the van der Waals equation of state is that due to 
C. Dieterici (published in 1899). The Dieterici treatment uses an excluded 
volume b as does the van der Waals equation. Its approach to the pressure 
defect due to intermolecular attraction is, however, different: it is, in some 
ways, more sophisticated. The van der Waals treatment emphasizes the 
diminution in momentum transfer because of the reduction in velocity of 
impact, whereas Dieterici emphasizes the reduction in number of molecules 
hitting the surface in unit time. The number of molecules to reach the wall in 
unit time with a specified velocity is less than the ideal number, because they 
need to possess a higher energy in order to escape from their neighbours. 
The energy needed to escape is proportional to the density of molecules «, 
so we may write 

number = ideal number x exp 


where is a suitable constant. Clearly n is proportional to 1/K, so that fin 
may be written as at/ V where a is another constant. The observed pressure P 
may then be written 

have fin 


P = /'ide.i exp | 
Since P Mm i V^ = RTvm have finally 

P(V-b)=* RTcxp(-^—\. (6.16) 

105 Imperfect gases 

It will be noted that apart from Clausius' relation all these equations involve 
only two arbitrary constants. As we shall see later this makes it possible to 
deduce a law of corresponding states for gases obeying the van der Waals, 
Berthelot or Dieterici equation. 

6-2-4 Attraction of the walls 

We have spent some space in discussing intermolecular attraction as the cause 
of a pressure 'defect' in real gases. What of the attraction between the 
molecules and the walls of the container? Does the pressure exerted by a gas 
depend on the nature of the container? Will the gas pressure be higher for a 
polar container than for an inert one? The answer is simple. The wall can have 
no net effect. Attraction may increase the impulse during the collision but the 
impulse will be reduced by an exactly equal amount due to the extraction of 
momentum from the wall. The net effect will be zero. 

range of inter-molecular 

range of surface forces 

Figure 33. Behaviour of a gas molecule as it escapes at A from the attraction of its 
neighbours and then approaches at B to within the range of action of the surface 
forces of the walls of the container. 

This is shown schematically in figure 33. The velocity of a molecule in 
the bulk of the gas is, say, c. When it escapes from the attraction of its neigh- 
bours its velocity has fallen to c' (position B). At B it comes under the attrac- 
tive forces of the wall. Consider the interaction with a wall molecule X of 
mass M. Suppose X is initially at rest. The molecule is accelerated towards X 
but also draws X out of the wall to X'. When collision occurs the velocity of 
the molecule (mass m) has been increased to c" whilst the velocity of the 
wall molecule is, say, v. The gain in momentum of the molecule m (c"— c") 
is exactly equal to the gain in momentum in the opposite direction Mv of the 
wall molecule. Thus the wall molecule is given an increased blow because 
of the attractive forces, but the increased momentum is exactly nullified by the 
momentum withdrawn from the wall. The net momentum transfer is the same 

106 Gases, liquids and solids 

as if no attraction had existed. On rebound the molecule leaves with increased 
velocity so imparting a larger momentum to the wall molecule, but as the 
gas molecule returns to position B' the attractive forces decrease its velocity 
to its original value c'\ at the same time they decrease the effective impulse 
on M, and it returns to the wall with the same momentum as though no 
attractive force had existed. The total momentum change is thus the same as 
if the wall exerted no force on the gas molecule. (This is simply a description 
of Newton's laws of momentum in terms of molecular interactions.) By the 
time the gas molecule reaches A' its velocity is back to c and the bulk behaviour 
of the gas now resembles that of a perfect gas. 

The problem discussed in this section is often avoided in elementary texts. 
Usually there is a bland statement that, by applying Newton's laws of 
momentum, the molecular momentum is reversed after collision with the wall. 
This is, of course, perfectly correct but it is equivalent to assuming that the 
wall exists at ZZ. It ignores the detailed interaction which takes place in the 
region between ZZ and the wall molecules themselves. 

6'2-5 Van der Waals equation and the Boyle temperature 

If we plot isothermals of PV against P we note that the Boyle temperature 
occurs when the slope of the curve is zero, at P — »■ 0. We therefore use the 

van der Waals equation, determine — (PV), and put it equal to zero at 

P = 0. 
From the original equation (6.13) we have 

P= XT a 

v-b v 2 ' 

so that py =£!¥—-. (6.17) 

V-b V 


— (PV)= IRT RT , + —z) — I 

dP \ V-b (V-b) 2 V 2 j\epJ T 

\ RTb a\/dV\ _,„ x 

If we specify that, at T = T B , this is zero when P = 0, this is equivalent to 
considering V as approaching infinity so that (V—b) is indistinguishable 
from V. From equation (6.18) we therefore have 

-RT B b+a = 0, 
or t.**-§-. . (6.19) 

107 Imperfect gases 

At all temperatures above T B , PV will increase with increasing P, so that 
there will be no inflexion in the PV curves. At T B itself we may substitute 

T = T B = — - in the van der Waals equation. Multiplying out equation (6.13) 

we have 

On the left-hand side we may, if the pressures are small, replace — by — -, 
and neglect the fourth term. We are left with 

PV+~-bP = RT. 

From our value of the Boyle temperature ( T B = — | we see that the sec- 
ond and third terms on the left-hand side are equal and opposite. Hence at T B 

PV = RT B 
for a real van der Waals gas. 

6-2-6 Van der Waals equation and the critical points 

The van der Waals equation is a cubic in V. If we plot P against Kwe obtain 
a family of curves shown schematically in figures 34 and 35. Let us consider 
a typical isotherm XDBACEY in figure 34 below the critical temperature. 
Join D and E by an isobar which cuts the curve at A. 

Figure 34. Stylized behaviour of a van der Waals gas. 
108 Gases, liquids and solids 

Figure 35. The van der Waals isotherms at various temperatures. 

Consider now the behaviour at A. Suppose we started with a homo- 


geneous phase. Along BC the slope — is positive. This means that if a small 


increase in volume occurs this is accompanied by an increase in gas pressure. 
There is thus a spontaneous expansion towards C. Similarly a small decrease 
in volume would lead to a spontaneous contraction towards B. These 
processes constitute the initial separation into two phases, gaseous at C liquid 
at B. The part BD is unstable, the liquid can remain in this region (region of 
superheating), but generally the phase spreads to D. Similarly CE is unstable, 
the gas can remain in this region of super cooling but generally the phase 
spreads to E.* In passing we may note that some isotherms can give negative 
pressures i.e. B may be below the Kaxis. This only occurs in the liquid phase 
and as we shall see this provides a means of estimating the tensile strength 
of a liquid. 

The two turning points B and C coalesce into a single region at Z on the 
critical isotherm (figure 35). We thus have two conditions determining the 
critical isotherm: 


for turning points such as B and C, 

/ 01 P\ 

I -r^i I = for the coalescence of these at the point of inflexion Z. 


♦ A thermodynamic argument quoted in J. K. Roberts (1960), Heat and Thermodyn- 
amics, 5th edn, Wiley, shows that the line EAD should cut the curve at such a position 
that the areas DBA and ACE are equal. The reasoning is as follows. In carrying the gas 
through a reversible cycle DBACED (i.e. from D back to D) the gas is restored to its 
initial state so that there is no change in entropy. Since the whole process is at a constant 
temperature this implies that the net work done is zero. The work done y given by the 
PdV areas in each part so that the area DBA must be equal to the area ACE. 

109 Imperfect gases 

Multiplying out the van der Waals equation we have 

PV+--Pb— —, = RT. 
V V 2 

Differentiating with respect to V at constant temperature we have 

P+Vt — ) --b\ — +— - = 0. (6.21) 

\8VJ V 1 \8V) T V 3 

Inserting the condition that for all turning points j — I — we obtain 

\oVJ T 

_ a 2ab ., „. 

P = —i t . (6.22) 

V 2 V 3 

This is the equation of the curve BKZLC (figure 35). The tip of this curve 

/ 8P\ 
corresponds to Z. This is where | — I of equation (6.22) is zero. We have 


2a 6ab „ ., „. 

— + —- = 0. (6.23) 

Y 3 v 

\dvJ T 

We call the volume at which this occurs the critical volume V c . From equation 
(6.23) we obtain 

V e = 3b. (6.24) 

Substituting in equation (6.22) we obtain for the critical pressure 

P c = -^-r • (6.25) 

27b 2 

Substituting V e and P e in the van der Waals equation we find the critical 
temperature T c from the relation 


(V c -b) = RT C . 

This gives 

T c = — L . (6.26) 


The nature of the van der Waals equation is such that all isotherms below 
T e have two turning-points; no isothermals above T c have even a single 
point of inflexion. We may note that empirically the boiling point (°K.) of a 
liquid at atmospheric pressure is about \T C . There is no simple theory for this. 

110 Gases, liquids and solids 

6-2-7 Magnitudes ofb and a in the van der Waals equation 

•2-7-1 Meaning of b. We have already seen that for a van der Waals gas b is four 
times the volume of the molecules treating them as spheres of molecular 
diameter a. In a gram-molecule therefore 

b = Wofr( a ^=^No\ (6.27) 

2) 3 

From a study of the compressibility of a gas we can find the best values 
of b to fit the data and hence calculate a using equation (6.27). Some typical 
results are given in Table 16; they are in good agreement with values of a 
deduced from diffusion, viscosity, etc. 

Table 16 Values of a deduced from b 



H 2 

N 2 

co 2 

so 2 







Meaning of a. The pressure of a gas on a container is due to molecular 
collisions. If the velocity component normal to the container wall is u, then 
as we saw in Chapter 4 we may say : 

Number of collisions cm. -2 sec. -1 = — 


Momentum change collision -1 = 2mu. (6.28) 

The rate of change of momentum cm. -2 sec. -1 which is the product of 
these is the ideal gas pressure 

-Pideai = nmu 2 = $nmc 2 . (6.29) 

In escaping from the attractive force of the last few neighbouring molecules 
the collision velocity is reduced from u to w— Ah, so that the momentum 
transfer is reduced to 2»j(h— Ah). The number of collisions cm. -2 sec." 1 
which we may call the 'flux' is still nu/2. At first sight it might seem that this 

should also be diminished to -(«— Ah). This however is not so. Although 


the molecules are retarded they are not, in this model, prevented from reaching 
the wall, so continuity in flux from the bulk of the gas to the wall is maintained. 
The flux does, of course, depend on the molecular velocity but near the wall 
where retardation occurs there is a slight 'crowding-up' of molecules, that is, 
a compensating increase in density of the gas. In this way flux-continuity is 

111 Imperfect gases 

maintained. This model should be contrasted with the Dieterici treatment 
described above. The observed pressure P therefore becomes 

nu . 

P — — 2»i(m— Am) = nmu 2 




An a 

Clearly P ideal — is the van der Waals term — - . We may write this as 
u V* 




= M.imu 2 )n. 

Now A(£mu 2 ) is the loss in K.E. per molecule in the u direction and n = — . 

Consequently the van der Waals term 

— = — (loss of K.E. per molecule) (6.31) 

or a = NqVQoss of K.E. per molecule) (632) 

assuming no changes occur on the average in the v ox w directions. 

We could now assume some law of force between the molecules and 
integrate for all molecules and all distances, extending from the position of 
the last collision near the wall to all the molecules throughout the bulk of the 
gas. Indeed the van der Waals treatment is basically concerned with relatively 
long-range forces. However, as we are interested only in an order of magnitude 
calculation we can arrive at a reasonable value of a by considering only the 
nearest neighbours. Because of the simple way this is done it actually over- 
estimates the effect of the attractive forces. The method is illustrated in 
figure 36. Every molecule within the hemispherical shell bounded by r^ = all 
and r 2 = 3<r/2 is a neighbour which actually touches the molecule considered. 


2 Figure 36. The range of attraction of surrounding gas 
molecules as a single gas molecule collides with the 

112 Gases, liquids and solids 

The work done in dragging the molecule away from these will be this number 
of neighbours multiplied by Ae. The volume of the shell is ifwfrj— rf)= 

-13a' = 13p m where v m is the theoretical volume of a single molecule. The 

number of neighbours is then n x 13t> m so that the energy lost in pulling away 

the molecule, i.e. its loss of K.E. in the « direction is 

Qmu 2 ) = n x 13 x v m (As). 
The van der Waals term is then 

a = N Vnl3v m (Ae). (6.33) 

Since b = 4N v m and N = Vn, 

we have a = — JV 6(Ae). (6.34) 

Inserting this in our value for T n 

„ 8a 8 13 1 26 Ar Ae 26 Ae 

r « = M = »4 "^JS = 27 "°* = 27T 

i.e. kT e as Ae. (6.35) 

This extremely crude model gives a reasonable value for a and explains why 
the critical temperature is nearly equal to Ae/k. It must be emphasized that 
this is an 'order of magnitude' calculation and that the final constant in front 
of Ae (ff a 1), is largely fortuitous. The model itself, however, is basically 

6-3 Some properties of the critical point 

6-3 • 1 Critical volume 

V c = 3b = 12 x volume of molecules themselves. Hence the mean separation 
between the molecules at the critical point is 

1 = (12)*ff ~ 2-4ff. (6\3tf) 

There is just about enough room between molecules to squeeze another. 


Mean free path 

Using our simple derivation, 

, 1 

A" J. 

X_ 1 _ 1 
a nno* 6tw„' 

113 Imperfect gases 

But nv m is the volume occupied by the molecules in 1 c.c. of space. At the 
critical point this is ^ c.c. Hence 

k_ 1 
a 6x-& 

= 2. (6.37) 

Thus a, X and the mean molecular separation are all roughly equal at the 
critical point. For ideal gases at ordinary temperatures and pressures the 
relevant quantities are roughly 3, 1000, 30 A. respectively. 

6-3-3 Critical coefficient 

The ratio — - is known as the critical coefficient. For all equations of state 


which involve only two arbitrary constants the critical coefficient is an explicit 
dimensionless number. The values obtained for the Berthelot, Dieterici and 
van der Waals equations are summarized in Table 17. 

Table 17 Critical points and critical coefficients 

Equation of V P T RTc 

state ' P C V C 

Berthelot RT C a 8a 4 

see equation (6.15) 26 9TJ> 2 TIRb 3 ~ 




9T c b 2 


Dieterici a a _e^ ^ 

see equation (6.16) 4e 2 b 2 4Rb 2 ~ 

Van der Waals ,. a 8a „ 

3/> ^ ^ 2*7 

see equation (6.13) 27b 2 21 Rb 3 

Experimental values for the critical coefficient are as follows: 

Gas He H 2 N 2 C0 2 Water acid 


31 30 3-4 3-5 4-5 50 

P C V C 

114 Gases, liquids and solids 


We see that, in general, for a simple gas —-— lies between 3 and 3-5 if the 


attractive forces are of a van der Waals nature. For this particular parameter 
the Dieterici equation gives better agreement than the van der Waals equation. 
For larger and more polar molecules the critical coeffiicent is generally 
greater than 4-5. 

6-4 Law of corresponding states 

An interesting result is obtained if we express volumes, pressures and tempera- 
tures in terms of their critical values. We put 

T P V 

6 = -, * = -, * = -. (6.38) 

-* e " c ' c 

Then the van der Waals equation becomes 

L+^\(34-l) = 89. (6.39) 

We note that the van der Waals constants a and b have disappeared in this 
reduced equation. This is, at first sight, surprising. We know that in general 
we need 5 parameters to describe the behaviour of a real gas, e.g. 

P = RT, V,a,b). 

However in using reduced parameters we are really feeding in two extra 

(8V\ /8 2 V\ 

— 1=0 and j — - J = 0, on the critical isotherm, 

so that two of the parameters are eliminated and we are left with 

7t = f{0, ft. 

This, in turn, implies that if two of the quantities, say 6, (j>, are the same for 
two substances the third parameter n will also be the same for the two 
substances. The assumption that the reduced equation is equally applicable 
to all gases is known as 'the law of corresponding states'. It implies that by a 
simple extension or contraction of the scales on which P and V are plotted 
all gases will give identical isotherms at the appropriate corresponding 
temperature. This conclusion holds quite well for gases of a single species, 
for example, monatomic or homonuclear diatomic gases, but for larger, 
more complicated molecules the agreement is poorer. 

It may be noted that any equation of state which involves only two arbitrary 
constants can be converted into a reduced equation of state containing no 
arbitrary constants. For example the Dieterici equation becomes 

*[2<S- 1] = exp ]2 (l - - j 1 . (6.40) 

115 Imperfect gases 



Expansion of gases 

Reversible adiabatic expansion of a real gas 

For an ideal gas adiabatic expansion gives a temperature drop given by 


For a real gas the temperature drop is greater because of the energy expended 
in overcoming intermolecular attraction. 

6'5-2 Joule expansion into a vacuum 

For an ideal gas the internal energy is independent of the volume so that there 
is no net temperature change. For a real gas work is done solely against the 
intermolecular attractions. We may consider this as work done against the 
internal pressure a/V 2 , and write it as 

fV=r± d V=a\±-±-], 
Jv,V 2 [V, V 2 \ 

where V t is the initial and V 2 the final volume. 


6-5*3 Joule-Kelvin expansion 

This is a type of expansion suggested by Lord Kelvin and carried out by him 
in collaboration with Joule in 1853. The purpose was to study (under more 


-high T 




porous plug 

■low T 



Figure 37. The Joule-Kelvin expansion, (a) The gas is driven at constant pressure 
Pi through a porous plug and withdrawn at a lower pressure P 2 , (b) isotherms 
showing that for an expansion from a higher to a lower pressure, the change in 
PV is positive at low temperatures and negative at high temperatures. 

116 Gases, liquids and solids 

carefully controlled conditions than Joule's earlier experiments) the heat 
changes occurring when a gas undergoes free expansion. The general principle 
is illustrated in figure 37(a). Gas in the left-hand part of the tube is driven at a 
constant pressure P t through an orifice or porous plug and withdrawn on the 
right-hand part at a constant, lower, pressure P 2 . Joule and Kelvin studied the 
temperature changes occurring after a steady state had been reached for a 
system that was thermally insulated. For the purpose of our treatment, 
however, we shall assume that heat sources or sinks can be applied so that the 
temperature T on both sides is maintained constant. 

Let Vi be the volume per gram-molecule on the left-hand side, V 2 the 
volume on the right-hand side. During the passage of 1 gram-molecule of gas 

External work done on the gas on l.h.s. = P t V x 

External work done by the gas on r.h.s. = P 2 V 2 

Hence net work done by the gas = P 2 V 2 -P 1 V 1 . (6.43) 

There is also work done by the gas against the internal forces. We have 
already calculated this in equation (6.42). The amount is 

a a 

vTv,- <**> 

Hence the total work done by the gas when one gram-molecule flows is 

If there is to be no cooling of the gas, this amount of heat has to be supplied. 
We may calculate the amount of heat involved assuming the gas obeys the 
van der Waals equation. We start with 


(V-b) = RT, 

and multiply out. Neglecting the term in — we have 

V 2 

PV=RT~+bP, (6 , 

or PV~ = RT~+bP. (6.46) 


In the second term on the r.h.s. we replace V by — , so that 

PV-~ = RT-P^+bP. (647) 

117 Imperfect gases 

If T is to be maintained constant 

Since Pi > P 2 the heat supplied to maintain T constant must be positive if 

— - b is oositive, i.e. if — > b. This means that if heat is not supplied there 

will be cooling. 

Experiments show that with nearly all gases there is a cooling at room 
temperature after a Joule-Kelvin expansion. With hydrogen, however, there 
is a rise in temperature unless the gas is first cooled to about - 80°C. 

If the system is thermally isolated we may estimate the temperature 
difference produced during a Joule-Kelvin expansion in the following way. 
We write equation (6.47) more fully as 

( pv ~vj 2 ~ { PV ~vji = RT *- RTlHPl - P2) (f?~*) • (6,49) 

We cannot complete this rigorously without the use of the second law of 
thermodynamics. It turns out that the l.h.s., which is the net work done by the 
gas, is equal to CVAI", where AT is the temperature drop T t -T 2 . Equation 
(6.49) becomes 

C K AT= -MT+(i\-P2) (fj.-*) • 
Since C r = C r -R we have finally for the cooling 

We see that there is a critical temperature below which there is a cooling, 
above which a heating of the gas. This inversion temperature occurs when 

*L = b 

or Tt = -j| = 2T B . (6.51) 

6*54 The reason for cat inversion temperature 

It is instructive to consider separately the two main terms involved in the 
cooling of a gas subjected to a Joule-Kelvin expansion. If the gas expands 
from Pi to P 2 , i.e. if Pi is greater than P 2 the work done by the gas which may 
cause cooling is 

118 Gases, liquids and solids 

P*Vi-PiV 1 + 

U yj 

=A[PK]f + work against internal attraction. (6.52) 

For a real gas in which there is molecular attraction the term on the right 
must be positive and must therefore always lead to cooling. This is not 
necessarily true of the first term. At low temperatures, in moving from Pj. to 
P 2 i the quantity A[FF]f , is positive [see figure 37(6)]. Consequently additional 
work must be done by the gas in expanding and this causes increased cooling. 
At high temperatures, however, well above the Boyle isotherm, &\PV]\ is 
negative, i.e. work is done on the gas as it expands and this opposes the 
cooling effect of the second term in equation (6.52). At some intermediate 
temperature the two effects just balance and there is neither heating nor 
cooling. This is the inversion temperature T t . 

Although this is helpful in showing how the inversion temperature arises 
it is, in some ways, artificial since it attempts to separate the 'imperfect' 
properties of the gas into two independent characteristics. As the van der 
Waals equation shows these two characteristics are, in fact, linked. 

6 - 5*5 An isoenthalpic expansion in the Joule-Kelvin experiment 

When a gas is expanded reversibly through nozzles or a porous plug under 
conditions of thermal insulation we have by the first law of thermodynamics 

dQ = dU+PdV 

0= U 2 -U l +P 2 V 1 -P 1 V 1 , 

so that 

U 2 +P 2 V 2 = U 1 +P 1 V 1 

H 2 = Hi. 

The porous-plug expansion is thus an isoenthalpic process. The temperature 
change associated with the pressure change must therefore be expressed 

in the form ( — I . For a small change in H we have 

— (SMS)* 

If we are specifying that H is constant dH = and we are left with 


The first term on the left is C P . Consequently 


119 Imperfect gases 

In the next step we have to make use of thermodynamic functions based on 
the second law, viz: 


The Joule-Kelvin temperature change is therefore given by 


/BT\ 1 -(?E\ \ 8T 'p 

\8P) B ~ C P \8P) T ~ C P 

Ar=i-Ap[r(^) r r]. (6.56) 

Using equation (6.45a) a little manipulation shows that the term T 

in equation (6.56) is equal to b. Thus equation (6.56) becomes identical 


with the relation given in equation (6.50) for the Joule-Kelvin cooling. 

Consequently the inversion temperature is again given by 

T, = — . 

The above treatment suggests that there is a single inversion temperature 
which is independent of pressure. This results from assuming that the pressures 

are low, so that — can be neglected. If the higher terms are included it may 

be shown that for the van der Waals equation of state, T t is a parabolic 

function of the pressure - indeed at pressures above — the gas expansion 


shows inversion at all temperatures, that is at high pressures it will heat up on 


This is illustrated schematically in figure 38 (taken from F. W. Sears 
(1953), Thermodynamics, The Kinetic Theory of Gases, and Statistical Mech- 
anics, 2nd edn, Addison-Wesley) where curves of constant enthalpy have 
been plotted as a function of temperature and pressure. In expanding from say 
point D to point E the temperature of the gas will rise - whereas an expansion 
from point A or point B to point C will produce a drop in temperature. 

In effect the temperature T t given in equation (6.51) for a low pressure 
expansion is the highest value of the inversion temperature. Some approxi- 
mate values of T, at low pressures are given in Table 18 and compared with 

120 Gases, liquids and solids 

Figure 38. Curves of constant enthalpy as a function of temperature and pressure 
(from F. W. Sears (1953). Thermodynamics, The Kinetic Theory of Gases, and 
Statistical Mechanics, 2nd edn, Addison-Wesley). 

Table 18 Inversion temperatures T t of some 
gases compared with 2T B (°K.) 


T B 







H 2 




N 2 




121 Imperfect gases 

Chapter 7 
The solid state 

We pass at once from gases to solids. In gases the atoms and/or molecules 
are almost completely free; in solids they are almost completely lacking in 
mobility. Indeed in a solid the thermal motion is so greatly reduced that 
individual atoms can move from their fixed positions only with the greatest 
difficulty. It is this which imparts to solids their most characteristic macro- 
scopic property - they maintain whatever shape they are given, they have 
appreciable stiffness. Although the atoms are in fixed locations they possess 
some thermal energy - in some cases individual atoms can diffuse through the 
solid but in general the process is extremely slow. Their main thermal 
exercise consists in vibrating about an equilibrium position. 

7-1 Types of solids 

There are three main types of solids: 

7 • 1 • 1 Crystalline solids 

The main feature here is long-range order. The molecules or atoms are in 
regular array over extended regions within each individual grain or crystal. 
With large single crystals the regular array may extend over an enormous 
number of atoms or molecules. For example with a single crystal of copper 
of side 1 cm., along any one direction there will be 30,000,000 copper atoms 
in regular array, the arrangement of the last few being (in the ideal case) in 
perfect step with the first. 

In spite of this extraordinary degree of regularity the individual atoms can 
vibrate over relatively large amplitudes about their equilibrium positions. 
Thermal vibrations can easily exceed one tenth of the mean spacing; yet the 
mean spacing is precise and determinate to an extremely high degree of 

7-1-2 Glasses 

Here the individual atoms or molecules are in ordered array over a short 
range but there is no long-range order. They are indeed, in structure, very 
much like an instantaneous snap-shot of a liquid and are sometimes called 
supercooled liquids. However molecular movement, a characteristic feature 
of liquids, is minute and the attempt to describe their flow properties in terms 

122 Gases, liquids and solids 

of an equivalent viscosity leads to astronomical values of the viscosity. For 
example from the viscosity of certain silicate glasses at elevated temperatures 
it is possible to extrapolate and deduce a room temperature value of the order 
of 10 70 poise. 

An American physicist has pointed out that this is equivalent to stating 
that if the glass were in the form of a rod of length equal to the circumference 
of the Universe and was subjected to half its breaking tension for a period of 
time equal to the age of the Universe, its total extension could amount to less 
than one thousandth of the diameter of an electron. [F. W. Preston (1942), 
Journal of Applied Physics, vol. 13, p. 623.] For such materials, room tem- 
perature viscosity is clearly not a very useful physical concept. 

Glasses may also be described as amorphous solids. 

7 • 1 -3 Rubber or polymers 

These consist of long organic molecules held together either by chemical 
cross links, by hydrogen bonds or simply by van der Waals forces. Below 
a critical temperature (the glass transition temperature) they behave very much 
like glasses, but with a greater degree of ductility; above this temperature they 
are generally rubber-like (see section 8-4 below). 

7-2 Main types of bonding in crystalline solids 

We summarize here four main types of bonding and the corresponding 
crystalline states. 


Van der 




Solid H 2 , Kr, 

Crystalline state 

Close packing of weakly 
attracted units. 


Strong NaCl crystal 

Giant aggregates of positive 
and negative ions clearly 
packed in a way consistent 
with neutrality of charge. 

Covalent Strong Diamond, Si, Ge 

Giant molecules with di- 
rected bonds, packing deter- 
mined by valency number 
and valency directions. 

123 The solid state 

Metallic Strong 


Metal atoms give up their 
valency electrons leaving 
metal ions in a sea of elec- 
trons. Forces between ions 
and electrons are central 
giving close packing. Strong 
attraction gives strength. 
Mobile electrons give con- 

7-3 Solid-liquid transitions 

If we heat a solid, the molecules or atoms acquire sufficient energy to escape 
from their lattice sites and ultimately melt, the temperature remaining con- 
stant until the whole of the specimen is molten. The quantity of heat required 
to convert a unit mass of the solid at its melting point to a liquid is called the 
latent heat of fusion. 

The melting point of a solid depends on the pressure to which it is subjected. 
For a solid which is more dense than the liquid (this is the usual case) the 
melting point increases with pressure: for a solid which is less dense (e.g. ice) 
the melting point decreases with pressure [see figure 39(a) and (6)]. If we 






Figure 39. Melting curves (solid-liquid equilibrium) as a function of pressure for 
(a) a solid more dense than the liquid, (6) a solid less dense than the liquid. 

combine this with the liquid-vapour transition we obtain a phase diagram 
as shown in figure 40. We note the following characteristics: 

124 Gases, liquids and solids 



Figure 40. Equilibrium curves for solid, liquid and vapour. O is the triple point 
where solid, liquid and vapour are in equilibrium. The liquid phase does not persist 
above the critical temperature. 

(a) for the liquid-vapour phase equilibrium the curve ends at C. Above 
this point which corresponds to the critical temperature T c the liquid 
ceases to exist; 

(b) by contrast for the solid-liquid phases there is no evidence that there 
is an end point to OY above which only one phase can exist. 

7-4 Consequences of interatomic forces in solids 

We describe here a few of the properties of solids which are directly explicable 
in terms of interatomic forces. 

7-4'l Heat of sublimation 

If we could take a van der Waals solid such as solid argon at absolute zero, 
and heat it until all the constituent atoms evaporate we should have pumped 
into the solid sufficient energy to break the bonds between all the atoms. This 
is the theoretical heat of sublimation L s . It is equal to the latent heat of melting 
(L P ) plus the latent heat of vaporization (L y ) plus a specific heat term involved 
in raising the temperature from absolute zero to the temperature of vapori- 
zation. If there were an exact energy balance the released atoms would just 
leave the solid with a minute escape velocity; in practice they will have con- 
siderable thermal energy. Thus precise comparisons between bond strengths 
and energy of sublimation is difficult. However as an order of magnitude 
calculation such correlation is easy, as we shall now show. 

If the potential energy between each atom (or each molecule) and its 
neighbour is Ae, and if each atom (or molecule) has n nearest neighbours, 

125 The solid state 

the energy required to break the bonds in a gram-atom (or molecule) will be 

L s = iN n\e. (7.1) 

This is allowing only for nearest-neighbour interactions, where N is Avo- 
gadro's number and the factor i is introduced to avoid counting each bond 
twice. For close packing, such as in a face-centred cubic structure or a hexa- 
gonal close-packed structure, n is 12. 

L s = 6AfiJV . (7.2) 

If As is in ergs and we wish to express L s in calories per gram-atom (or 
molecule) we obtain 

86xl0 15 Ae. 


Some typical results are given in Table 19, the *observed' values of L s being 
simply the sum of L F and L v . The values of Ae are obtained either from 
theoretical calculations of the van der Waals forces, or from the behaviour 
of the substance in the gaseous state (for example, the deviation from idea 
gas behaviour at high pressures). 

Table 19 Heats of sublimation 


As erg 

L s cal. 


1 or g-molecule' 1 




0-9xl0" ls 




4-8xl0~ ls 




16-5 xlO" 5 




25xl0- 15 



H 2 * 

4xl0~ ls 



N 3 * 

13xl0" 15 



* Not face-centred cubic structure. 

Apart from helium, where the effects of 'zero-point energy' are important, 
the agreement is very satisfactory. A more detailed consideration of lattice 
energy, latent heat of sublimation and intermolecular forces will be given 
in the next chapter. At this stage we merely draw attention to the direct 
agreement with the simplest types of solids. We may also note that the 
latent heat of fusion is generally much lower than the latent heat of vapori- 

126 Gases, liquids and solids 

zation. From the point of view of thermal energy the liquid state, near the 
melting point, is much nearer to the solid state than it is to the gaseous state. 

7-4-2 Surface energy 

The atoms in the bulk of a solid are subjected to the attraction of atoms 
all round them, but at the free surface they are subjected only to the inward 
pull of the atoms within the solid. Consequently the surface layers have a 
higher energy than those in the bulk and this excess is called the free surface 
energy y. 


3 Jr 

Figure 41. Face-centred cubic crystal of side 2a. (a) The surface atom X has 8 
nearest neighbours at B, B, B, B and C, C, C, C ; whereas in the bulk each atom has 
1 2 nearest neighbours, (b) If the crystal is pulled apart in the plane PQRS each 
atom such as X breaks 4 bonds. 

We may calculate this in terms of the bond energy for, say, a typical face- 
centred cubic solid where all atoms except those in the surface have 12 
nearest neighbours. An atom X in the surface has only 8 nearest neighbours. 
This is shown in figure 41(a). If the crystal contains a total of N atoms and 
the surface contains N, atoms the total bond energy of the crystal, counting 
only nearest-neighbour interactions, is then 


= iAe{12iV-4iV J }. (7.4) 

(This assumes that there is no change in lattice spacing in the surface layers, 
a point that is still in some dispute.) 

If there were no surface atoms the energy would be iAex 12N. Thus the 
energy is reduced by i&e x 4N,. Since Ae is negative this is a positive increase 
in energy which is indeed the surface energy. If the surface area of the crystal 
is A, 

yA »iAfix4JV f . (7-5) 

127 The solid state 

Wo may obtain this result more simply by considering a rectangular bar of 
crystal of cross-sectional area A. Any section across the crystal contains 
N, atoms, so if we pull the crystal in two each surface atom breaks four bonds 
and the work done per atom is 4Ae [see figure 41(6)]. Hence the total work 
done is 4Ae x N„ But the area exposed is 2A. 

2yA = 4AnxN.. (7.5a) 

This is identical with equation (7.5). If in the crystal orientation we have 
considered there are z atoms per sq. cm. we see from equations (7.5) and 
(7.5a) that 

y = 2xAexz. (7.6) 

But the energy necessary to sublime z atoms from the bulk of the crystal 
[see equation (7 J)] is 

/, = 6Asz. (7.7) 

Hence the surface energy is equal to one third of the heat of sublimation 
of a single atomic layer of atoms. For different orientations the fraction 
will be a little different but generally it will be of the order of $ to \. For a 
face-centred cubic lattice of side 2a as shown, the volume occupied by each 
atom is 2a 3 . If M is the molecular weight and p the density 

= 2a 3 or 

Nop 2N p 

For the (100) face that we have considered the area occupied by each atom 
is 2a 2 . The number of atoms z per sq. cm. is — — . Hence 


This of course refers to a specific face and to a specific structure. For other 
faces and structures we should merely change the numerical factor by a small 
amount. A reasonable average value for the factor in front of equation 
(7.9) is 0-3. This relation should apply best to van der Waals solids such as 
solid neon, argon and krypton. However the surface energy of such materials 
is not known. We have therefore applied the model to metals, the results being 
given in Table 20. The quantitative agreement between the calculated and 
observed values is poor, but the agreement in relative values is surprisingly 

128 Gases, liquids and solids 

good. If all the calculated values are multiplied by a correction factor of 0-4 
the calculated values all agree with observation to within about 10 per cent. 
The basis for this correction factor is discussed in the next chapter. 

Table 20 Surface energies of some metals in solid state 


M.P. "C. 

L,kcal. mole' 1 
at 25°C. 

M 3 

y erg cm.~ 2 

Calc. Obs* 






1,300 ~450 






1,200 ~500 






3,300 1,300 






4,000 2,000 






5,000 2,100 






6,700 2,900 

* Observed measurements are generally at elevated temperatures so that they are some- 
what smaller than the values to be expected at 2S°C. These values are from M. C. Inman 
and H. R. Tipler (1963), 'Interfacial energy and composition in metals and alloys', 
Metallurgical Reviews, vol. 8, 1963, pp. 105-66. 

With liquids, the surface energy is relatively easy to determine; with solids, 
particularly ductile solids, it is far more difficult. The most direct method 
with metals is to hang a wire of uniform cross-section in a chamber at constant 
temperature and observe the region where necking just begins. (Such experi- 
ments give meaningful results only at temperatures approaching the melting 
point.) If necking occurs at a height h above the free end it implies that 

Weight of wire of length h = 2nry, 

where r is the radius of the wire. For most metals y turns out to be of the order 
of 500-2000 erg cm." 2 

With a 'perfectly brittle' solid like mica the surface energy can be deter- 
mined by finding the force required to open a crack. If we consider a cleaved 
specimen as shown in figure 42(6) we may allow the force F to move and to 
open up the crack as shown. The work done = JjJ Fdy, this is expended in 
changing the elastic strain energy in the two mica leaves and also in creating 
two new surfaces each of area A. The work done can be measured, the elastic 
energy calculated and the difference may then be put equal to 2yA. For 
mica in a good vacuum y has a value of about 5000 erg cm. -2 ; in air about 
300 erg cm. -2 In general such experiments cannot be carried out satisfactorily 
with ductile solids such as metals or polymers, since the plastic work absorbed 

129 The solid state 




Figure 42. Determination of surface energy y of solids (a) by the necking of a 
specimen at section XX; this occurs when the weight of the wire of length 
h = 2nry (b) by bifurcation of a mica strip. The work done in opening up the strip 
is equal to the change in elastic strain energy plus the increase in surface energy 
(2jv4) involved in creating 2 surfaces of area A. 

by deformation at the tip of the crack may be several hundred times greater 
than the surface energy. For example, values of y often come out to be of the 
order of 10 s erg cm. -2 Therefore the experiments can be carried out only on 
brittle solids. Reasonable values have been obtained on glass (A. A. Griffith 
1920) and as mentioned above on mica (J. W. Obreimov 1931, A. I. Bailey 
1955). Recently Gilman has shown that the technique can be applied success- 
fully to many crystalline solids if the experiments are carried out at liquid air 
temperature - he showed that at this temperature the ductility is 'frozen 
out*. Surface energies of the order of a few hundred erg cm. -2 are obtained 
for materials such as NaCl, LiF. 

We may mention here a simple feature involved in the surface energy of 
solids which differentiates it markedly from that of liquids. Suppose we have 
a surface of area A possessing surface energy y. If we increase the area by a 
small increment the work done per unit increase may be written 

d(Ay) 8y 



For liquids the term — - - is zero, since, on account of atomic mobility, the 

structure of a liquid surface is unchanged when the surface area is increased. 

This is not so with solids; when the surface is stretched (even only notionally) 

130 Gases, liquids and solids 


the atoms are pulled apart and y diminishes so that — is negative. This 


implies that with solids the surface energy is not the same as the surface force 

(more strictly the line tension in the surface; see Chapter 10 on the surface 

tension of liquids); with liquids they are identical. 

Although one cannot equate surface energy with surface force there is 
generally some tension in the free surface. Even if it is as large as an effective 
line-tension of 1000 dyne cm. -1 , this is minute compared with the elastic 
modulus of the solid, usually about 10 12 dyne cm.- 2 Clearly forces of this 
magnitude can have very little visual effect on metals; so they do not influence 
the shape of a solid in the same way as they do liquids. A centimetre cube of 
copper is not dragged into spherical shape at room temperature because of 
surface tension. However, with very minute particles some such effects may 
be observed. 

Surface energy of solids plays a very important part in sintering; and it is 
a crucial factor in the wetting of solids by liquids. From the point of view 
of atomic mechanisms it is worth emphasizing that the same forces which 
produce the pressure defect in a real gas, are responsible for the surface energy 
of a solid and the surface tension of a liquid. 

Elastic stiffness 

If atoms or molecules in a solid are subjected to a stress and thereby displaced 
from their equilibrium positions the interatomic forces will tend to restore 
them to their original positions. If the distortions are not too large, when the 
stress is removed the atoms will (under ideal conditions) return to exactly the 
same situation as before; the elastic modulus is defined as the ratio of the 
applied stress to the deformation produced. This will be discussed in detail 
in a later chapter. Here we merely quote the result for Young's modulus which 
is defined as 

Tensile stress increment Sa 

E = 

increase in length Ax 
original length x 

= x^. C7.ll) 


The elastic modulus of a solid may be easily demonstrated if one considers 
the force-displacement curve. Figure 43 shows the force between two neigh- 
bouring atoms as a function of separation. At low temperatures where thermal 
energy can be neglected, the equilibrium separation occurs where the attractive 
force equals the repulsive force. Thus OA is the value of the equilibrium 

separation x. The slope of the line ZAY through A is — . Hence the quantity 


131 The solid state 






f A 



I (repulsion 



Figure 43. (a) Force-separation diagram illustrating the decrease in elastic modu- 
lus with expansion, (b) Potential energy-separation diagram illustrating the in- 
crease in separation which occurs when the temperature is increased (thermal 

x — is the exact one-dimensional analogue of Young's modulus as defined 


If the solid is heated thermal expansion (see below) shifts the equilibrium 

position from A to A'. The slope — is seen to be smaller and this decrease 


is appreciably greater than the corresponding increase in the new equilibrium 
separation O'A'. Thus the effective modulus is reduced. If we tried to construct 
a simple cubic structure out of these pairs of atoms the number in any square 

cm. of an atomic plane would be — - so that the stress (as distinct from the 

x 2 

F dx ldF 

force) is — and the strain is — . Then Young's modulus becomes .In this 

x 2 x xdx 

case we see that the slight increase in x, with rise in temperature, augments the 

decrease in — so that both -terms of the modulus are reduced by a rise in 

temperature. However as already mentioned the change in x is in general small 

compared with the change in — since F is usually a 'strong' power function 


of x. 

We conclude that if the behaviour of the solid as a whole resembles that 

of the diatomic model discussed above the elastic modulus decreases as the 

132 Gases, liquids and solids 

temperature is increased. This is true of practically all solids except rubber- 
like solids. 

7-4-4 Thermal expansion 

If the length of a solid specimen at 0°C. is l a and at f°C. is /, one can usually 

/, = /o(l+«X 

where oc is the coefficient of linear expansion and for most solids is of order 

io- Io c.-' 

This behaviour may again be explained in terms of the forces between the 
individual atoms. It is most convenient to plot the potential energy curve as a 
function of separation. If we do this for two atoms we obtain the typical 
curve shown in figure 43(6). At absolute zero the equilibrium position 
(ignoring zero-point energy) is at X, so the equilibrium separation is YX. 
As the temperature is raised the energy rises to, say, Y' and one of the atoms 
will oscillate relative to the other between positions F and Q' about a mean 
position X'. Because of asymmetry in the shape of the potential energy curve 
the length Y'X' is greater than YX. Thus the mean separation increases with 
temperature along the curve XX'X* .... 

If an assembly of atoms in a solid behaved similarly this would explain 
thermal expansion. Of course there are added complications. In a solid one 
ought to treat this problem in terms of the free energy; this introduces an 
additional term but it is generally small compared with the effect described 
above. Again in the diatomic model the atoms dissociate when the energy 
reaches 0. This is equivalent to sublimation of the solid and tells us nothing 
of the melting process that occurs somewhere en route. This is because 
melting is a co-operative process involving many atoms and a satisfactory 
model must be multi-dimensional to allow for their behaviour (see Chapter 
10). Finally we may note that increased temperature may give rise to trans- 
verse oscillations. In some types of crystals this may actually lead to a 
contraction in one crystallographic direction as the temperature is raised. 

Nevertheless the simple model given above provides a reasonably satis- 
factory explanation of thermal expansion in solids. In a later chapter we shall 
indeed use it to derive an explicit relation for the thermal expansion in terms 
of intermolecular forces. We shall see that the results are in quantitative 
agreement with experiment. In general the potential energy-separation curve 
provides a very useful way of approaching many of the physical and mech- 
anical properties of solids. 

133 The solid state 

Chapter 8 

The elastic properties of solids 

If we subject a solid rod to tension we pull the atoms further apart; if we 
compress it we push them closer together. The interatomic forces resist 
these changes and, if the distortions are not too large, the body will return 
to its original dimensions when the external forces are removed; the deforma- 
tions are then called elastic. 

8-1 Some basic elastic properties 

8-1-1 Elastic moduli 

If a solid rod of uniform cross-sectional area A is pulled with a tensile force 
F the tensile stress is denned as 

ffSS A' 






Figure 44. (a) Deformation in tension, (b) deformation in shear, (c) deformation 
under hydrostatic pressure. 

If, as a result, the rod increases its length from /to /+A/ the fractional 

increase — is called the linear strain e [see figure 44(a)]. Within the elastic 

range Hooke showed that the stress is proportional to the strain. The ratio is 
denned as Young's modulus E: 

134 Gases, liquids and solids 


Linear stress A. 
- ~ Linear strain ~ "a7" * * ' 


A solid may also be deformed in shear. For example if we consider a rect- 
angular element as in figure 44(b), we may apply a shear stress t (force -r area) 
across the faces AB and CD. This must be accompanied by equal shear stresses 
t along CB and AD so that the net couple is zero tsee Chapter 9, section 912). 
The element is then distorted through a small angle which is termed the 
shear strain. The strain is again found to be proportional to the stress and the 
ratio is termed the rigidity modulus n: 

" = |- (8.2) 

A solid may also be deformed by a hydrostatic pressure p. If the solid has 
volume v, and the change in volume is — Av, then the volume strain is - Av/v 
and this is found to be proportional to p. The ratio is the bulk modulus K. 


The reciprocal of this modulus is known as the compressibility, /?. 

Finally when a body is extended in tension to give a linear strain «,, it 
contracts in directions at right angles to the direction of tension. The 
fractional contraction e, in the transverse direction is found to be propor- 
tional to si. The ratio 

Contractile strain e, _ . . 

— — ; — = — = Poisson s ratio v. (8.4) 

Linear strain e t 

For most solids v has a value between i and J and this implies a decrease in 
volume under tension. If v = i (as is nearly the case with rubber) there is no 
volume change in tension or compression. 

There are several relationships between E, n, K and v which we shall not 
prove here. For example 

n = 2irb) «> 

■■'■■3(i^- (8 -«> 

We see from equation (8.6) that for values of v = 0-25 to 0-33, JSTis comparable 
with E; for a value of v = \, X= oo. This implies that the material is 
incompressible, which is what would be expected since tension and com- 
pression produce no volume change in such a material. 

136 The elastic properties of solids 

8>1 '2 Torsional rigidity 

If a circular tube of internal radius r u external radius r 2 , length / is held at 
one end and subjected to couple r at the other end, every part of the tube is 
subjected to shear. Suppose the free end is twisted through an angle ^. 
The ratio TI4> is known as the torsional rigidity. We shall not derive it but 
simply quote the relationship of T/# to be: 

where n is the rigidity modulus. 

8-1-3 The bending of a cantilever 

If a bar is encastered at one end and deflected sideways by a force F at the 
other end, the deflection produced depends on F, on the section of the bar, 
its length and its Young's modulus. Although we shall not derive the equation 
here, we shall indicate the principle of the analysis involved. Suppose figure 
45(a) represents, in exaggerated form, the bending of a beam. The sections 
p, q, r, s, t indicate how the beam is deformed. We make a notional cut 
through the bar at section r ; then the material to the left of r is subjected to a 
bending moment Fx. If the beam is to be in equilibrium this must be counter- 
balanced by a reverse moment of the same magnitude. The material above 
AB is stretched and so exerts a tensile force to the left. The material below 
AB is compressed and so exerts a resisting force to the right. The beam bends 
to just that point where these two forces produce a couple exactly equal to 
Fx. This implies that between the stretched and compressed portions there 
is a line AB which is unchanged in length. AB is, in fact, a part of the line 
XX' which represents the original length of the unstressed bar; it is known 
as the neutral axis. We may note in passing that the section between s and t 
has to carry the smallest bending moment and is therefore distorted least: 
the end part of the beam is indeed straight. On the other hand the bending 
moment is a maximum at CD; here the curvature is a maximum. This raises 
an interesting point which is usually ignored by physicists and engineers. 
The encastered portion CDEF is considered completely rigid and subjected 
to zero bending moment. This implies that at the boundary CD the bending 
moment suddenly drops from a maximum value on the right to zero on the 
left. Such a situation is impossible, for there must be a transitional region 
HK within the encastered zone where the bending moment gradually falls 
to zero. In practice this implies a certain amount of slip between the bar and 
the clamp at ED and FC or some other type of yieldi ng in the clamped zone. 
This is not of general importance in calculating the deflection of a loaded bar. 
It can however be of very great importance; for example in the propagation 
of a crack or in the splitting of mica described earlier as a means of deter* 

136 Gases, liquids and solids 

(c) If 

\ife^.^^/ : 'jfe»-> 1 ...fe, ., 



Figure 45. (a) Elastic bending of a beam encastered at one end and subjected to a 
force F at the other. The line XX' is the neutral axis where there is neither extension 
nor compression of the original length. (b) The bending moment has a minimum 
at the free end and a maximum at the point where the beam enters the encastered 
zone. Simple theory assumes that it then drops at once to zero; in practice there 
must be some transition region HK. (c) The bifurcation region of a mica speci- 
men resembles the encastered zone of a bent beam. 

mining the surface energy. The conditions at the region of bifurcation 
resemble those of the encastered region in a bent beam. The analysis of the 
forces in this region is very difficult indeed. 

If the moment of inertia of the section of the beam about the vertical axis 
is / and the length of the beam is / the deflection y produced at the end by a 

137 The elastic properties of solids 

force Fis given by 
Fl 3 


El 3 

For a beam of rectangular section, thickness a, width b, this becomes 

_F 4^ 
y ~ E*ab 3 ' 



8-1-4 Young's modulus in terms of interatomic force constant 

In the previous chapter we showed how the elastic modulus is related to the 
force displacement curve. We shall now make this more quantitative. Consider 
a solid composed of a single species of atom and let us assume a simple cubic 
structure, the separation between each atom being a. Let the bar have unit 
cross-sectional area and consider three neighbouring planes I, II and III 
(figure 46). We assume that there are only nearest-neighbour interactions; 



po qo 







o— \i 







Figure 46. Simple cubic array of atoms, (a) undeformed, (b) when a force f is 
applied to each atom in plane III all the planes undergo an increased separation x, 
(c) a single force f acts in a similar way on a single atom q. 

then if an atom q in plane H is displaced in a direction normal to the plane, it 
experiences a restoring force only from atom p in plane I and atom r in 
plane III. If a tensile force/is applied to atom r the spacing between the atoms 
in that row will each be increased by a small distance x (where x < a). The 
restoring force which each atom in this row experiences will be proportional 
to x. Let the restoring force be kx where k is, in fact, a measure of the slope 
of the force-displacement curve shown in the previous chapter [figure 43(a)]. 

138 Gases, liquids and solids 

Since there are 1/c 2 atoms per sq. cm., the force per sq. cm. (the stress) to 

/ kx 
achieve a separation of the planes of this amount will be — = — - . This dis- 

ar tr 


placement is equivalent to a linear strain of amount - . Then we may write for 


Young's modulus 


_ Tensile stress a 2 k „.„,.* 

E = — — = _= -. (8.10) 

Strain x a 


8-2 Propagation of longitudinal waves along an elastic bar 

We extend this to a very simple description of wave propagation. Our major 
(and weakest) assumption is that we may isolate the interaction of one atom 
with its next neighbour. Suppose we hold planes I and III fixed [see figure 
46(c)] and displace atom q relative to atoms p and r by an amount x. Then 
p exerts an attractive force kx on q and r exerts a repulsive force of the same 
magnitude. The resulting restoring force is 2kx. If the atomic mass is m the 
equation of motion is 

mx+2kx = 0. 

This gives simple harmonic motion of period 



Suppose now we give an impulse to the free end of the bar and displace the 
row of atoms in plane I to the right (figure 46). These then act on the atoms in 
row II which in turn are pushed to the right and so act on the atoms in plane 
III. In this way the impulse is propagated through the solid. The time taken 
for each plane to respond to the push of the neighbouring plane is of the 
order of one quarter of the vibrational period T. Thus the impulse jumps 
from one plane to the next in a time 

In Im In lm 
t =i — /— ~ — /— . (8.12) 

4 *J 2k 5-7 V k 

For our purpose we may take 2n/5 -7 to be unity. The disturbance thus travels a 

distance a in time /— . The velocity of propagation is therefore 

t *J m 

or vm /(--). (8.13) 

a/ \maj 

139 The elastic properties of solids 

m k 

We recognize — as the density p, and the quantity - as Young's modulus E. 
<r a 





This is the same as the equation for the velocity of a longitudinal wave in a 
bar in terms of its macroscopic properties. It shows that the vibrating atom 
is the 'messenger' for wave propagation. 
The frequency of vibration is 

2na \ P ) 


For, say, iron we have £= 2xl0 12 dynes cm. -2 , p = 7-8 g. cm. -3 and 
a ^ 2-5 x 10"* cm. Then v a 5x 10 12 vibrations per sec. This is the right 
order of magnitude. 

The main defect in this model is that the vibrations of the atoms are not 
isolated but are coupled. A rigorous treatment does not, however, sub- 
stantially alter the nature of this analysis. 

8-3 Bulk moduli 

8*3-1 Bulk modulus of an ionic solid 

We consider here an ionic solid of the NaCl type in which the Na + and Cl - 
ions are distributed on a cubic lattice as shown in figure 47. We consider a 
single ion and its coulombic interaction with all the charged ions around it. 
As the figure shows there are : 

6 neighbours of opposite sign at distance x 
12 neighbours of same sign at distance \^2x 
8 neighbours of opposite sign at distance \/3x, etc. 

If each ion carries charge e (or -e) the potential energy of one ion in 
relation to all its neighbours is 

-e* T^ 12 8 1 

= — [21 . . .]. (8.16) 


Proper summation to infinity shows that the correct value, known as the 
Madelung constant A, is 1*75 for the rock-salt structure. 

140 Gases, liquids and solids 



Figure 47. The structure of the NaCI crystal, (a) General view, (b) 6 neighbours 
of opposite sign at distance x. (c) 1 2 neighbours of same sign at distance J2x. 
(</) 8 neighbours of opposite sign at distance V3x. 

Assuming a repulsive potential energy term between each ion pair of the 
form A\x 9 , the summation of all the repulsion terms will give Bjx 9 where B 
is an appropriate constant. The potential energy of one ion allowing for both 
attraction and repulsion is 

e 2 B 

-1-75- + 4' 
x x 9 


If there are N positive and N negative ions per mole the potential energy 
for a mole of NaCI is 

U = K2ATo«) = No(-175 j+^) 

where the \ is introduced so that we do not count each bond twice. 
The equilibrium spacing a occurs when U is a minimum. 


Putting — at x = a equal to zero we find that B has the value 


Equation (8.18) becomes 
r/=-l-75AT c 2 

P— 1 
[x 9x 9 } 


l-75c J a 8 


The next step appears complicated but is basically an attempt to connect the 
internal energy with an elastic modulus, in this case the bulk modulus K. If we 
apply a hydrostatic pressure p this produces a volume change — dV, and the 
work done by the external forces on the crystal is then — pdV. The ions are 
squeezed closer together and if the whole of this work goes solely in increasing 
that part of U which arises from interatomic forces, we have -pdV = dU or 

p sr - 




K~ - 




8V 2 




141 The elastic properties of solids 

We need to express this in terms of the ionic spacing x. We have 


e 2 u _ _£_ Ibu\ _ j_ /eu dx\ _ a_ leu dx\ 
ev 2 ~ av \av) ~ dv\ dx*dv) ~ e x \8x *dv) 

'(dx\ 2 8U/d 2 x\ ,„„„ v 

{d-v) + Tx[dV 2 ) + -- C8 - 22) 

8 2 U/dx] 

ex 2 ' 


At the equilibrium separation (x = a) — = 0; consequently the second 


term vanishes so that 


We now relate Kto the atomic spacing x. For a gram-molecule 
V = 2NoX 3 . 
dx 1 

dV 6N0X 2 


To obtain K of equation (8.23) we differentiate equation (8.19) twice to obtain 

8 2 U (dx\ 2 

— - and use equation (8.24) for { — I . We obtain 
8x 2 \dVJ 

At equilibrium when x = a and V = 2N a 3 we find 

J8: = ™2 X £! = 0.78^. (8.26) 

9 a* cr 

For rock salt, assuming complete charge separation on the ions, e is the 
electronic charge, 4-8 x 10" 10 e.s.u. The equilibrium spacing a = 2-82x 10 -8 
cm. Substituting in equation (8.26) we obtain 

K= 2-8x10" dyne cm."\ 

whereas the experimental value is about 2-4 x 10 11 dyne cm. -2 Taking into 
account the basic simplicity of the model this is surprisingly good agreement. 
It implies, of course, that the bonding is fully ionic. In most 'ionic' crystals only 
part of the bonding is ionic, the rest is covalent. 

We may also express K in terms of the bond energy U of the lattice in its 
equilibrium state. Substituting x = a in equation (8.19) we have 

14-0 te 2 \ 
142 Gases, liquids and solids 

Hence from equation (8.26) 

J5:= ^f! = Jl^ = ^2, (8.28) 

9 a 4 2M>« 3 K 

where Vis the volume of a gram-molecule. We shall see that a similar result is 
obtained for van der Waals solids. 

It is interesting to note that, for a free NaCl molecule in the vapour state, 
the distance between the Na + and CI" ions is 2-36 A. compared with 2-82 A. 
in the crystal lattice. This indicates the important part played by the surround- 
ing ions of opposite sign in opening up the ionic spacing. It is relatively easy 
to make a rough quantitative estimate of this. For a single ionic pair in 
the free molecule we may write 

M = Z£? + 4. (8.29) 

r r 9 

The equilibrium spacing r occurs when — = 0. This gives the value of 

A = -£-. Hence 

In the crystal the coulombic interaction of the ionic charges gives alternative 

positive and negative energies; as we saw above the final amount for each ion 

is - 1 -75 e 2 \r where 1 -75 is the Madelung constant. The repulsion term always 

involves a positive energy; however the potential falls off so rapidly with 

distance that we need only consider the 6 nearest neighbours at distance r for 

6A 6r a e 2 
which the repulsion energy is — or — ~ . The energy per ion is then 

f 9r 

The minimum occurs when — = 0, i.e., when 

1-75 _ 6rg 
r 2 r 10 ' 

r 8 = 3-43r§ or r = 117r . (8.32) 

If r is taken as 2-36 A. this would give for r a value of 2-76 A. which is close 
to the observed value. Even for a one-dimensional infinite chain molecule 
NaClNaClNaCl . . ., the ionic spacing is already increased from 2-36 A. 
to 2-69 A. 

143 The elastic properties of solids 

8-3-2 Bulk modulus of a van der Wads solid 

We consider a simple van der Waals solid. If it consists of single atoms or 
spherical or nearly spherical molecules it generally acquires a close-packed 
(face-centred cubic) structure. If the molecules are long chain hydrocarbons 
the packing tends to be very nearly hexagonal with the chains lying parallel 
to one another. We shall discuss only the face-centred cubic type of structure 
such as is found in solid argon, neon and krypton. For a pair of atoms at a 
distance x apart the van der Waals attractive forces produce a potential energy 
proportional to x~ 6 ; the repulsion produces a potential energy which appears 
to be proportional to x~ 12 , rather than to x~ 9 as in the previous ionic model. 

Figure 48. The structure of a face-centred cubic structure of a solid inert gas. 
Each atom has 12 neighbours at distance >/2x apart, 6 neighbours at distance 
2x apart, etc. 

The potential energy between two atoms may be written 

x 6 x 




Each atom has 12 nearest neighbours distance y/2x apart and 6 neighbours 
2x apart, etc. (see Figure 48). The contribution from the attraction forces is 

f 12^ A 

1 _ l ' 6A 

The convergence is so rapid that we may ignore further terms. The repulsion 
term may be grouped in a single constant C, so that the resultant potential 
energy may be written 

«= -1-6— +— . 
x 6 x i2 


144 Gases, liquids and solids 

For No atoms the total energy will be $N times this. As before we may differ- 
entiate U to find the equilibrium condition when x - a. 

U = -0«NoA (±-\ £) . (8.35) 

At the equilibrium separation 

0,--0-4^. (8-36) 

The volume of the gram-atom for this structure is 

V=2N„x 3 , (8.37) 

so that as before 

— = — — , (8.38) 

dV 6N0X 2 

Repeating the previous procedure we find for the bulk modulus 

JST=1^. (839) 

a 9 

The nearest-neighbour distance of atoms is y/2x and this is almost identical 
with the equilibrium separation r for a single pair of atoms. We may thus 
rewrite equation (839) as 

K =11 X ¥L (8.40) 

a 3 r% 

The second factor in equation (8.40) has a very simple significance. By differ- 
entiating equation (8.33) we may determine B in terms of the equilibrium r 
of one atom in relation to its neighbour. We shall then find the potential 

energy As of an isolated pair of atoms is simply - — . Consequently from 

equation (8.40) 

*=^. (8.41) 

For the inert gases and for the smaller molecules such as H 3 and N 2 , As is of 
order (5 to 25) x 10 -ls erg, and a is about 3 x 10" 8 cm. in the solid state. 
We obtain 

K = (4 to 20) x 10 9 dyne cm."* (8.42) 

This is about one hundred times smaller than for ionic solids. 

146 The elastic properties of solids 

We may also express K in terms of U . Combining equations (8.37) and 
(8.34) we have 

K=-4^-=-S^. „ (8.43) 

Na 3 V v ' 

This is similar to the result for ionic crystals except that the numerical factor 
is appreciably larger. 

8 -3 -3 Bond energy, or lattice energy, and heat of sublimation 

Before going on to discuss the bulk modulus of metals we make a few general 
comments on the bond (or lattice) energy and the heat of sublimation. 

The total bond energy U of a lattice (the lattice energy) is a fundamental 
quantity deriving from the forces between the constituent atoms and ions. 
It is not easily open to direct measurement. The nearest physical quantity is 
the heat of sublimation L s , which in practice is a little greater than the sum 
of the latent heat of melting L F and the latent heat of vaporization L r .We may 
easily form an estimate of L s for an ionic crystal. When sodium chloride is 
evaporated, the process does not involve pulling the positive and negative ions 
apart to infinity; this would be the true lattice energy. Vaporization generally 
produces individual NaCl molecules. L s is thus less than U by the value of 
the interaction energy between Na + and Cl~ in the individual NaCl molecule. 
From equation (8.31) the lattice energy per ion in the solid state is obtained 
by putting r = 1 -\lr where r is the ionic spacing in the crystal, r Q the spacing 
in the free NaCl molecule. The value is 

e 2 

"lattice = (1-33), 


and for the gram-molecule 

r/ = -^f!(l-33). 

The energy per ion pair in the free molecule is obtained from equation (8.30) 
by putting r = r . We obtain for the free NaCl molecule 

- _ e * fi\ 

^(molecular — I _ I t 

'o \V 

and for the gram-molecule 

_ Ne* (8\ 

"molecular- ~ [gj • 

The molar heat of sublimation is therefore 

Ne 1 
L s = U - Molecular = — (0"44) « - \U . (8.44) 


146 Gases, liquids and solids i 

The latent heat of sublimation, which we can measure with a fair degree of 
accuracy, is thus about one third of the lattice energy. Tables of values support 
this conclusion. 

With a van der Waals solid sublimation merely separates all the molecules 
(or atoms) as individual molecules (or atoms). The heat of sublimation is thus 
a fairly accurate measure of the bond energy. We may easily form an estimate 
of this from the above analysis. 

From equation (8.43) we have 

and from equation (8.41), this becomes 

Uo = 3 ' 2AeK . (8.44a) 

a 3 

But from equation (8.37) we see that when the crystal is at its equilibrium 

separation (x = a), V= 2N<>a 3 . Substituting in equation (8.44a) we have 

Uo = 6-4N Ae. (S.44h) 

This is the same as the value obtained in equation (7.2) except that the 
numerical factor is slightly different. This arises from the different assump- 
tions made in the two models. 

Metals present a very special problem. They consist of metal ions in a sea 
of free electrons. It is not possible to describe the forces between the ions and 
electrons in terms of a simple power law. Consequently the analysis we have 
given above is not possible. A reasonably good estimate of the bond energy 
U can be made, but the treatment is very difficult and involves concepts which 
we shall not discuss in this book. Even if we knew the bond energy it would not 
correspond to the latent heat of sublimation. The bond energy would cor- 
respond to the work done in separating all the ions and all the free electrons 
to infinity; whereas in the vapour state the metal exists essentially as free 
atoms. Thus U = L S + ionization energy of the metal atoms. A similar type 
of difficulty exists with covalent solids. 

8-3-4 Bulk modulus of metals 

For the reasons given above it is not possible to derive the elastic properties 
of a metal using the simple type of analysis that we have applied to ionic 
and van der Waals solids. However we may form a reasonable estimate of the 
bulk modulus by noting that, in general, 

K=-c^, (845) 

where U is the total bond energy per gram-atom (or molecule), V the atomic 
(or molecular) volume and c a constant depending on the law of forces. For an 

147 The elastic properties of solids 

ionic crystal c = 1 ; for a van der Waals solid it is 8. For metals we can only 
guess at a value of c. We also have to recognize that we have no simple way 
of calculating U , but we can use the sublimation energy L s (which can be 
determined experimentally) as some measure of U . As a matter of interest 
we shall try the relation 

K=3- s 


where 3 is purely an empirical factor. This represents an attempt to allow 
both for the unknown law of force between metal ions in the lattice and for 
the discrepancy between the bond energy and the latent heat of sublimation. 
For L s we have simply added the latent heats of melting and of vaporization 
(taken from C. J. Smithells (1962), Metals Reference Book, 3rd edn, Butter- 
worth) and have ignored any specific heat terms. The results are given in 
Table 21. 

Table 21 Latent heat of sublimation L, and bulk modulus K of some metals 


It per g-atom 

Atomic vol. V 

Kx 10 u 

dynecm.~ x 


erg xlO 10 

cm.~ 3 



















































The agreement is surprisingly good in view of the over-simplifying assump- 
tions made. 

8-4 Elastic properties of robber 

The elastic properties of rubber are very different from those of most other 
solids such as ionic solids, covalent solids and metals. In tension or shear the 
modulus is 1000 to 10,000 times smaller. On the other hand Poisson's ratio 
is almost i, so the bulk modulus is relatively high; it approaches that of a 
liquid. The two most striking elastic features are (a) rubber has enormous 

148 Gases, liquids and solids 

extensibility - up to 400 per cent compared with perhaps 1 per cent for ordin- 
ary solids, (6) the elastic modulus increases with increasing temperature. This 
is the only solid known which shows such an effect. We recognize that these 
two properties are very similar to the behaviour of a gas in compression. 

Since rubber is known to consist of long chain organic molecules we 
realize at once that its enormous extensibility and low modulus cannot arise 
from the stretching or bending of the C — C bonds, since these are extremely 
strong and difficult to deform. The chains are, in fact, coiled up in higgledy- 
piggledy fashion and the result of applying a tension is to pull the chains into 
partial alignment. Thermal energy attempts to increase the disorder in the 
chain and therefore resists the aligning action of the tensile force. Conse- 
quently the modulus in tension increases with temperature. 

If the molecular chains were completely separate the material would be 
a very viscous liquid and indeed natural latex is such a liquid. In order to 
become a rubbery solid the molecular chains must be cross-linked at a 
relatively small number of points. We now consider briefly the properties of 
a typical segment lying between two cross-links. Suppose the C — C bond has a 
length A and there are N such bonds in a segment ; because rotation about the 
C—C bond, in contrast to stretching or bending, is extremely easy, as a first 
approximation we may ignore any stiffness which arises from this factor. 
The segment then consists of N freely pivoted bonds. We may then ask: what 
is the most probable length of a segment of N units each of length A? 

Assuming that each link is completely free to take up any position relative 
to its neighbour* we recognize this as the problem of a 'random walk' in 
three dimensions or a 'random flight'. 

We indicate briefly here how this may be calculated. Consider first a random 
walk in one dimension. Suppose a man takes N steps, each of length A, along 
a line; each step may be forwards or backwards. What is the probability that 
he will have travelled a distance x from his starting point? If out of N steps 
A are forwards and B are backwards 

A+B = N 

and AA-BA = x. (8.47) 

Hence A 


B = l\H-" x I • (8-48) 

Any way of taking TV steps such that A are forwards and B backwards will give 

* This ignores the requirement of fixed bond-angles between the carbon atoms. If this 
is taken into account, and rotations about the bonds are permitted, the value of /5 in 
equation (8.54) is merely changed by a small numerical factor. A more serious conceptual 
difficulty is that in the random walk any given step may be covered many times, whereas 
in a real molecule the location of a molecular link can only be occupied once. 

149 The elastic properties of solids 

the resultant distance x. The number of ways Win which this can be achieved 

For simplicity assume N to be even so that A and B [from equation (8.48)] 

/N\ N 
are integers. We use the Stirling approximation Nl = ( — I and take x < NX 


(in the rubber molecule this is equivalent to assuming that the distance 
between the two ends of the segment is very much less than the fully extended 
length of the segment). Then 

laW= Nln2-- 

2m 2 
or W = Bs' flx ' . (8.50) 

where PI = . The probability P(x) of the distance x occurring is directly 

Ally A 

proportional to W so that 

P(x) = Ce~ ffx \ (8.51) 

The constant C may be determined by observing that 

x=°NX /•<» 

£ PM = 1 ~ ¥{x)dx. (8.52) 

*«s-NA J -00 


This gives C = ^ . 

We now turn to the three-dimensional random walk. If the coordinates of 
the starting point are 0, 0, and those of the final point are x, y, 2 on 
orthogonal axes, the probability of finding the final position is 

w -» P 3 -fiH* I +y 2 +' 1 ) , aK ~ 

P(*> y, 2) = -je . (8.53) 

Normalizing for three dimensions it turns out that the /? 2 in equation (8.53) 
is 3 times & for the one-dimensional random walk, i.e. 

e> = \± (8-54) 

We are only interested in the exponential factor. We may therefore write for 
the most probable length of a segment 

P(r) = Ae~ p * r \ (8.55) 

150 Gases, liquids and solids 

If one end of the segment is fixed, the probability of finding the other end in 
an element of volume d x is 

Pdx = Aa 




We see that the probability density P is a maximum when r = 0, that is the 
segment makes its greatest effort to have zero distance between its ends. 
This is to be expected since if all steps In the random walk are equally probable 
a large number will contain as many forward as backward ones. This does 
not mean that the most probable distance between chain ends is zero. If we 
are interested only in the length r and not the direction then (as in our treat- 
ment of molecular velocities in a gas) we may consider one end of the segment 
fixed and determine the number of the other ends lying in a shell of volume 
4nr 2 dr. Then the number II with lengths between r and dr is 

Tldr = A47ir z e~ pr2 dr. 

Figure 49. (a) Probability density P as function of distance between the ends of 
a long chain assuming a three-dimensional 'random walk', (b) most probable 
distance r between the ends of a chain with N segments each of length X occurs 

for t m = - = i.*Ji\Ni), (c) random arrangement of a chain containing 1000 

segments, each segment set at the appropriate bond angle, but being given a 
random choice of 6 equally-spaced angular positions : this is a reasonably close 
approximation to complete randomness (from L. R. G. Treloar (1958), Physics of 
Rubber Elasticity, 2nd edn, Oxford University Press). 

151 The elastic properties of solids 

The result is shown in figure 49(b); the maximum occurs for a value of r 
given by 

rm = - = WdN). (8.57) 


A simple analogy is to consider the concentration of bees around a drop of 
honey. The greatest density occurs near the honey but the greatest number of 
bees may well be found between 5 and 6 cm. from the drop, rather than in the 
first cm. 

Reverting to equation (8.55) the probability density is P and this is related 
to the (configurational) entropy of the segment by the relation 

S = k log P = constant- kfi 2 r 2 . (8.58) 

If we apply a tensile force /to one end of the segment and extend it in the 
direction of r by an amount dr, the external work done is f dr. Assuming that 
this involves no volume change we may equate it to the change in the Helm- 
holtz free energy [equation (3.23)]. 

fdr = dA = d(U-TS). (8,59) 

For an isothermal extension 

dr Br 
= — + 2kTfi 2 r. (8.60) 

The term — - refers to the change in internal energy with extension, that 

is, with uncoiling of the segment by rotation about the C — C bond. As there is 

very little energy change (in our model no change at all) in this we may 

neglect this term and write 

/-2^-?. (8.61) 

We see at once that (a) /is proportional to r so that the segment has Hookean 
properties, (b) /is proportional to T, (c)/is proportional to fi 2 , that is to 1/JV: 
the smaller the number of bonds in the segment, the greater its resistance 
to elastic extension. 

So far we have only discussed the force that the segment exerts in resisting 
extension; whatever its length it is always attempting to contract. In bulk 
rubber which consists of a network of segments and chains it turns out that 
even if the rubber is subjected to uniaxial compression the overall effect, 
because of the large Poisson's ratio, is to extend the chains. The specimen 
therefore resists both extension and compression in a manner resembling 
the behaviour of a single segment discussed above. 

152 Gases, liquids and solids 

We see that the elasticity of rubber arises from the entropy or randomness 
of the chain segments. It is for this reason that it resembles the elasticity 
of an ideal gas in compression where the pressure may be considered as arising 
from the reduction in entropy associated with a reduction in volume. Indeed 
one notices the presence of the 'gas constant' R as a fundamental term in the 
expression for rubber elasticity in equation (8.61). In view of this one might 
well designate R as the 'rubber constant' in 'ideal rubber' elasticity. 

153 The elastic properties of solids 

Chapter 9 

The strength properties of 


9-1 Deformation properties 

9 • 1 • 1 Ductile properties 

The strength properties of solids are most simply illustrated by considering 
the behaviour of a homogeneous specimen of uniform cross-section subjected 



L I 


tensile stress 


P C 


section A 



linear strain 

Figure 50. (a) Tensile specimen and (£) typical stress-strain curve, showing 
elastic deformation along OA, plastic yielding at Y and work-hardening along YZ 
A brittle solid fails at a tensile stress S. 

to uniaxial tension [figure 50(a)]. If we plot the true stress a against the linear 
strain e (i.e., the fractional increase in length) we may obtain a curve as 
illustrated in figure 50(A). The portion OA represents elastic deformation. 
The strain is proportional to the stress and the deformation is reversible. If 
the material is ductile, elastic deformation will proceed until at some critical 
stress Fthe onset of permanent or plastic deformation occurs. If we continue 
along the plastic curve there is generally an increase in yield stress with 
deformation. This is known as work-hardening. If at some point B we reduce 
the stress the material recovers elastically along BO', where BO' is very nearly 
parallel to OA. The displacement OO' is the permanent plastic extension 

154 Gases, liquids and solids 

produced in the specimen. On reapplying the stress the deformation follows the 
curve OT3Z. 

Experiments show that under simple uniaxial compression a cylinder will 
first deform elastically and then yield plastically at the same compressive 
stress Y. 

Some insight into the yield criterion is provided by considering the effect 
of hydrostatic pressure, that is a stress situation in which the comprehensive 
stress on the material is the same in all directions. If we subject the specimen 
to a hydrostatic pressure P we find that plastic flow does not occur even if P 
exceeds Y. We must still apply a uniaxial stress (either in tension or com- 
pression) and its magnitude if plastic flow is to occur is still Y. Now analysis 
shows that the only part of a stress field which is unaffected by hydrostatic 
pressure is a shear stress. We conclude that plastic flow is associated with a 
critical shear stress. This is fully supported by microscopic studies which 
show that plastic deformation is always accompanied by slip of atomic 
planes over one another. 

9-1-2 Shear stress 

Consider a rectangular bar of uniform cross-sectional area A [figure 51(a)]. 
Suppose we apply a tensile force F, so that the stress is a = FjA. Consider a 
thin slice of material making an angle 6 to the direction of F. On one side of 


Figure 51. Shear stresses produced by tensile stresses: (a) vertical stress a, (b) 
horizontal stress <r, (c) two-dimensional hydrostatic stress a; this produces a 
resultant zero shear stress. 

the slice there is a force F cos 0, on the other side a force of equal magnitude 
in the opposite direction. These forces constitute a shear. The surface area 

155 The strength properties of solids 

of the slice is A/sin so that the shear stress is given by 

Fcos0 F „ . . 
ti = — - — = — cos 6 sin 
A A 

= a cos 9 sin 9. 
The maximum occurs for 9 = 45° and has a value r = - 


If we apply a tensile stress a to the faces be, da, the shear stress on the slice 


Tj = a cos ^ sin ^ 
= a sin cos 9. 


This stress is equal to ti and in the opposite direction. If the two systems of 
tensile stresses a are superposed (to constitute a two-dimensional hydrostatic 
tension), the two shear-stress systems completely annul one another. This is 
the basis for the statement that hydrostatic stresses do not in any way change 
the existing shear stresses in a system. 

Finally we note that the conventional method of representing a shear 
stress by two equal and opposite parallel stresses r is misleading [figure 52(a)]. 





F, = rA, 

Figure 52. Sketch showing that a shear stress must always involve a pair of equal 
orthogonal shear stresses. 

Such a stress system would produce a couple which would produce continuous 
rotation of the stressed element. There must be an opposing set of shear 
stresses s if static equilibrium is to be achieved. Consider a rectangular 
element length x, width y, depth z. On the upper and lower faces we apply 
shear stresses r; on the side faces shear stresses s. For equilibrium take 
moments about O. Shear stress t implies a shear force txxy; its couple is 
t x xy x z. Similarly shear stress s implies a couple sxyzxx. 

Then vxyxz = sxyzxx, 

t = s. (9.3) 

We see that a shear stress always involves a pair of equal orthogonal stresses. 
In practice, of course, one is often unaware of this. For example in shearing 
a long thin slice as in figure 52(c), the horizontal shear force ft » xA x may 

156 Gases, liquids and solids 

be very large and the vertical shearing force F 2 = rA 2 may be vanishingly 
small if A 2 is very small, although the value of the shear stress r must be 
the same in both cases. 

A tensile stress Y produces a maximum shear stress Y/2 at 45° to the direc- 
tion of Y. (The same applies to a uniaxial compressive stress Y.) For an 
isotropic material shear will therefore occur in slip directions at 45° to the 
direction of the applied stress. If the material is not isotropic shear may occur 
more easily in some directions than others; if the shear stress is exceeded in 
this direction slip will occur in these more favourably orientated directions. 
This is often observed with single crystals, and is shown schematically for a 
simple shear stress in figure 53. 

(a) (b) 


Figure 53. The effect of a shear stress on the yielding of a single crystal (a) when 
the shear plane is parallel to the shear, (b) when it is inclined. 

We may now explain the behaviour shown in figure 50(A). When the stress 
is first applied the atoms are displaced from their equilibrium positions; 
the resistance to deformation is determined by the interatomic forces. When 
the tensile stress reaches a critical value the shear stress is sufficient to 
produce slip along an appropriate plane and plastic yielding occurs. However 
the whole stress must still be supported by the displaced atoms. Consequently 
when the stress is removed there is elastic recovery and the modulus, which 
arises from the interatomic forces, is essentially the same as originally. 

9 • 1 -3 Indentation hardness of ductile solids 

Engineers and metallurgists often wish to determine the strength properties 
of their materials without going to the trouble of preparing tensile specimens 
and carrying out full-scale tensile tests. A very convenient way of doing this 
is to measure the indentation hardness. A very hard indenter (a hard steel 
sphere in the Brinell test, a diamond pyramid in the Vickers test) is pressed 
under a load JKinto the surface of the material to form a plastic indentation. 
When the indenter is removed the diameter of the indentation is measured 
and its area A determined. The mean pressure over the indentation is then 

P = j • (9.4) 

In the industrial test procedure A is the surface area of the indentation; 
from the point of view of making p physically meaningful, the projected area 

157 The strength properties of solids 

is more appropriate. The difference however is generally small. A study of 
the stress situation around the indenter shows that almost two-thirds of p is 
in the form of a hydrostatic component, and therefore plays no part in pro- 
ducing plastic flow. Consequently only one-third of p is active in producing 
the indentation. Thus as a first approximation 

P = 3Y, (9.5) 

where Y is the uniaxial yield stress of the material. This relation is well 
substantiated in practice. 

Further, if the material does not appreciably work-harden after yielding 
in tension its yield stress Y will be very nearly the maximum stress the 
material can support before it pulls apart, i.e., its ultimate tensile strength 
(u.t.s.). The latter is therefore about one-third the indentation hardness. 
Since p is normally measured in kg. mm. -2 and the u.t.s. in ton in. -2 the 
conversion factor must be divided by 1 -58. Consequently 

u.t.s. (ton in." 2 ) = - — p = 0-2lp (kg. mm." 2 ). (9.6) 


This conversion ratio is widely used for polycrystalline homogeneous 
materials and is reliable to within a few per cent. 

Typical values for the indentation hardness in kg. mm. -2 are 0-25 for 
krypton at -220°C, 0-8 for ice at - 10°C, 1 for indium, 4 for lead, 40 for 
polycrystalline copper, 120 for mild steel, 900 for ball-bearing steel, 2000 for 
sapphire and over 10,000 for diamond at room temperature. 

9-1 -4 Calculation of critical shear stress for single crystals 

We now consider the shear stress at which two neighbouring planes in a 
single crystal can be caused to slide over one another. Consider the arrange- 
ment in figure 54 for a typical face-centred cubic crystal and apply a shear 
stress t to the planes X and Y. If is the strain produced and G is the rigidity 
or shear modulus 

t= GO. 

Atom B in its initial site in the lattice is in a position of minimu m potential 
energy. As it is displaced the P.E. increases until a position of unstable 
equilibrium is reached at B' - the P.E. curve at this point is horizontal but is a 
maximum. The P.E. curve thus has the form shown in figure 54(6). The force 
curve is found by differentiating the P.E. curve: it is drawn in figure 54(c) and 
shows that the maximum force occurs when B has reached some intermediate 
position B" between B and B'. The angle of strain at this point is about \. 
If G remained constant we could say that B is sheared over to B' for a shear 
stress of magnitude 

*».--;• (9.7) 


158 Gases, liquids and solids 









Figure 54. Shearing of a perfect crystal, (a) Neighbouring planes of atoms at 
X and Y, (b) the potential energy curve as the top plane is slid over the bottom, 
(c) the equivalent force-displacement diagram, (</) a more realistic form of the 
force-displacement curve. 

The lattice could not resist a greater stress than this so that plane X would 
glide over plane Y for this value of t. 

This is an overestimate since it assumes that G is a constant and has the 
same value as for small strains. This is rather unrealistic since for larger 
distortions we expect the modulus to decrease. We may form a better estimate 
of r m by assuming that the shear stress is a sine function of the displacement. 
It must clearly be zero for displacements x = 0, x = a/2 and x = a, where a 
is the atomic spacing. We may thus use a function of the form 

t = A sin 



where A is a suitable constant. We may determine A by specifying that the 
initial slope near x = 0, where the strains are small, corresponds to the shear 
modulus G of the material. For small x 

2n In 

t = A sin — x a* Aa — x = GO, 
a a 


159 The strength properties of solids 

when 8 is the shear angle - . Consequently 

a 2n ^ x A G 

A — x = G - or A = — . 
a a In 


The shear stress reaches its maximum when the displacement is midway 
between B and B'; here sin ( — x ) is unity so that 


Tin = An = 



A stress of this value will carry atom B over to position B" and then for a 
slight increase in r it will flick over to B' and then to C. Thus slip along the 
atomic plane will take place for a shear stress of about G/6. A more realistic 
study of the atomic force field suggests a curve like that shown in figure 
54(d). This reduces the critical shear stress r m even further but no amount of 
adjustment can reduce it to a value less than about G/30. 

We conclude that the maximum shear stress r m the lattice can withstand 
is of the order of ^ to -^ G. Before discussing the relation of (T m ) theoretical 
to (r m ) experimental we may ask a very simple but pertinent question. Once 
we have dragged atom B to the top of the potential hill at B' [figure 54(b)] 
why does it not glide over the rest of the potential hills until planes X and Y 
have completely slid off one another? Is not the position analogous to a 
'frictionless' helter skelter, where once the carriage has reached its highest 
point it can proceed indefinitely over all hills which do not exceed that height ? 
The simple answer is that such a model would be appropriate if the atoms in 
row Y were absolutely rigid in space. Because they are themselves part of an 
elastic system the behaviour is much more like the interaction of two 
(frictionless) combs the teeth of one being dragged through those of the other 
[figure 55(a)] . Up to a certain tangential stress both sets of teeth are distorted 
reversibly [figure 55(6)]; beyond this point the teeth from the upper comb 

(a) (b) (c) 


Figure 55. The displacement of the teeth of two sets of engaging combs (a) 
initially, (b) after a small stress has been applied, (c) after the stress has been 
sufficient to cause slip : the elastic strain energy in the teeth is lost by vibration. 

160 Gases, liquids and solids 

escape from interaction with the lower comb and flick over into their new 
equilibrium position; whilst the teeth of the lower comb flick back into their 
original position [figure 55(c)]. The whole of the distortional energy has 
disappeared as vibration of the teeth. This has a close analogue in the slip of 
crystals. Practically the whole of the work of plastic deformation is dissipated 
as vibrational energy in the lattice, that is as heat. Not more than a few 
per cent of the energy is retained as strain energy in the lattice. This has an 
interesting corollary in an unexpected area. The friction of metals is largely 
due to adhesion, shearing and deformation within and around the regions 
of real contact; it is, in fact, a process which is generally dominated by plastic 
flow around the contact zones. The plastic work virtually all appears as heat. 
This is the reason for Joule's observation that in factional heating there is 
quantitative agreement between the work done against friction and the heat 
liberated. If large amounts of energy could be stored in the lattice this equival- 
ence would not be observed. 

9-2 Dislocations 

9-2-1 Observed shear stress, need for dislocations 

Experiments on pure single crystals show that minute shear stresses are 
sufficient to produce permanent, i.e. plastic, deformation. In general (r m ) 
observed is of the order of one thousand times smaller than (r m ) theoretical. 
The theoretical value cannot be wrong to this extent. This discrepancy has 
led to a search for sources of weakness in the lattice, such as flaws or 
imperfections: the simplest type is that known as the edge dislocation 
shown in figure 56(a). It involves, essentially, the insertion of an extra half- 






— r 



• 4 

i • 


















• i 

i • 

















• * 

» • 






















• i 


» • 














• « 

» • 












































m i 

» • 





Figure 56. (a) An edge dislocation, (b) the displacements which occur when a 
shear stress is applied and the dislocation at B moves out to E. 

161 The strength properties of solids 

plane of atoms AB into the lattice. If a shear stress is applied as shown, 
atom B flicks over and joins with atom F for a very small value of r. Then 
C joins up with G and D with H so that for a very small stress we have 
produced slip by one atomic spacing DE [figure 56(A)]. In effect what we are 
saying here is that the theoretical value of r m assumes that slip occurs 
simultaneously across the whole plane KE; in the dislocation model slip is 
able to occur one row of atoms at a time and this leads to an enormous 
reduction in shear stress. 

We can already see that there are two basic problems associated with the 
above model. First, how are the dislocations originally produced in the crystal ? 
Secondly when the edge dislocation has slipped out of the crystal as in figure 
56(6), the crystal has become perfect so that it has its ideal theoretical 
strength. As this is not generally observed there must be some process for the 
generation of further dislocations whilst slip is occurring. Both these questions 
are part of a very large subject which we shall not pursue further in detail. 
We mention however a few further points of interest. 

9-2-2 Direct experimental evidence for dislocations 

The first and most direct evidence for the existence of dislocations is due to 
Dr. J. W. Menter, who studied, with an electron microscope, very thin 
specimens of platinum phthalocyanine. This is a metallo-organic compound 
which forms well-defined crystals in which the platinum atoms lie on crystal 
planes 12 A. apart. In the electron microscope the organic part of the crystal 
scarcely absorbs electrons whilst the platinum atoms strongly absorb. In 
transmission, therefore, a microscope with a resolution of say 10 A. is able 
to resolve the planes of the platinum atoms. Typical election micrographs 
[see J. W. Menter (1956), Proceedings of the Royal Society, A236, p. 119, and 
F. P. Bowden and D. Tabor (1964), Friction and Lubrication of Solids, 
Clarendon Press, Plate III] reveal the presence of a defect strongly resembling a 
classical edge dislocation. In pure metals the atomic spacing is of the order 
of 3 A. and electron microscopes are not generally able to resolve distances 
smaller than about 5 A. Consequently it is not possible to 'see' dislocations in 
metals in the direct way that they are revealed in phthalocyanine. However, 
it is now possible to identify certain features revealed by the microscope as 
dislocations. This has led to a very extensive application of electron micro- 
scopy and diffraction to the study of dislocations in thin metallic films. 

9-2-3 Upper yield point 

The edge dislocation is a region of large elastic stress. At LBC in figure 56(a) 
the material is heavily compressed, at MF strongly extended. If impurity 
atoms are present in the lattice they will tend to concentrate at regions where 
they can relieve the stress. For example a small impurity atom will tend to 
take the place of the original atom B. This will lower the intensity of the stress 

162 Gases, liquids and solids 

field around the dislocation. It is precisely the presence of the stress field that 
makes it easy for a dislocation to move. The presence of the impurity at B 
thus tends to 'pin' the dislocation and a larger stress than normal will be 
required to make it move. Once the dislocation has moved across to position 
C or D it is in its normal environment and is now much easier to move. This 
explains the behaviour for example of certain mild steels where the dislocations 
are pinned by interstitial carbon or nitrogen. When such materials are tested 
in tension or compression or shear they show an 'upper yield point' (see 



Figure 57. Schematic diagram showing upper yield stress at B and a second 
upper yield at G which occurs after 'ageing*. 

figure 57, point B), then a region of fairly constant lower yield stress (CD) 
followed by a conventional work-hardening curve. If at some point F the 
stress is removed and the metal is 'aged' the carbon and nitrogen can diffuse 
to the dislocations and again pin them. The further deformation of the 
specimen may then show a second 'upper yield point* (G). 

9*2 -4 Work-hardening 

Dislocations interact with one another. Continued deformation produces 
interlocking of dislocations and makes it more difficult for them to move. 
Although the details are still the subject of discussion this is the basic reason 
for work-hardening as a result of deformation. 

9-2-5 Screw dislocations 

Apart from edge dislocations there are 'screw' dislocations as shown in 
figure 58(a). If we follow a sequence of atoms, say, in the top plane starting 

163 The strength properties of solids 

Figure 58. (a) Screw dislocation, (b) crystal growth on an edge providing two 
points of attachment at E and F, (c) crystal growth on the edge of a screw dis- 
location which provides a self-perpetuating growth step. 

from A and moving to B C D E F we end up at the point F which is an 
integral number of atomic spaces below A. This is essentially a screw move- 
ment. If the dislocation line XY moves towards CD the slipped region AXYZ 
increases in area. 

9-2-6 Crystal growth 

The presence of sites for nucleation is very important in crystal growth. 
On a perfectly smooth atomic plane an atom can find only one point of 
attachment. On a step, each atom can find two points of attachment [figure 
58(6)] . Growth will therefore occur very much more easily until the whole of 
the step ABCD is filled, when the surface is then plane. A screw dislocation, 
however, provides a self perpetuating growth step; as further atoms become 
attached to the step they build up a spiral 'staircase' and so provide further 
steps for further condensation. This mechanism accounts for the growth 
spiral that is often observed on the face of crystals, [see figure 58(c)] 

164 Gases, liquids and solids 

9-3 Brittle solids 

9-3-1 Brittle properties 

We now consider the behaviour of a solid which normally shows no ductility. 
If it is extended it stretches elastically and then snaps at some critical tensile 
stress S. If the material is homogeneous it will crack along a plane normal 
to the direction of the stress. If the specimen is subjected to a hydrostatic 
pressure P the tensile stress necessary to cause brittle failure is found to be 
P+S (figure 59). This implies that brittle failure occurs when the resulting 






/ I \ 


Figure 59. Brittle failure : (a) in pure tension the specimen fails for a tensile stress 
S, {b) when a hydrostatic pressure P is applied brittle failure occurs for a tensile 
stress of P+S. 

tensile stress across some appropriate plane exceeds a critical value, in this 
case S. In its idealized form this is the stress necessary to pull one plane of 
atoms completely away from a neighbouring plane. 

We may estimate the theoretical brittle strength in a number of ways. 
If the bar has unit cross-section and y is the surface energy, the act of snapping 
the specimen is to create two new unit areas of surface: the work done is 2y. 
If intermolecular forces are appreciable only over the distance of an atomic 
spacing, say 5 A., and the stress over this distance is assumed constant, 
we may write 

work done = Sx 5 x 10" 8 = 2y. 


For paraffin wax, where y is of the order of 100 erg cm." 2 this gives a value 
of S of about 40 x 10 8 dyne cm. -2 or 4000 kgf. cm. 2 For rock salt for which 
y is about 500 erg cm.~ 2 the theoretical value is of order 20,000 kgf. cm. 2 
This is enormously larger than the observed strength. 

1 65 The strength properties of solids 




,- f. 

~*~V>v .-•. •'. 






Figure 60. (a) Force-displacement curve for calculating the theoretical brittle 
strength. (/>) role of a crack, length /, tip radius p, in facilitating brittle failure, (c) the 
formation of a crack by the 'pile-up' of dislocations. 

We may also estimate the strength from the force-separation curve 

[figure 60(a)]. When the force reaches the point B it is great enough to pull 

the atoms apart. This occurs when — = 0. If we express the energy U in 

Bx k 

terms of x we have — = - F. Hence failure occurs when 

8x 2 

= 0. 


We have not derived an expression for the energy of a crystal in terms of 
the atomic separation in one dimension. We may however estimate the 
hydrostatic tension necessary to pull a crystal apart. For an ionic crystal 
we have (from the previous chapter) 

U = —const 

P— 1 
I* 9x»J 


Ignoring the constant and differentiating, 

Tx 1 ' 

x = 5*xa = l-23a. 

2 10a 8 „ 

r ii 


This shows that the crystal can be expanded by a hydrostatic tension until when 
the ionic separation has increased by 23 per cent , i.e., the volume has increased 
by 85 per cent, it will disintegrate. If the bulk modulus J? remained constant for 

166 Gases, liquids and solids 

these large strains it would imply a pressure equal to 0-85 K, but because of 
the large curvature of the force-separation curve in this region the stresses are 
appreciably smaller. Simple calculation shows that it should be reduced by a 
factor of about 3-5 so that the corresponding pressure is about 0-24 K. 
Using the theoretical value of K from the previous chapter, this is 7 x 10 10 
dyne cm. -2 or 70,000 kgf. cm. 2 This is of the same order as that deduced 
from the surface energy. In practice a hydrostatic tensile experiment would be 
difficult if not impossible to carry out; most estimates based on direct tension 
give a theoretical value for the brittle strength of the order of 

S:-, (9.16) 

where E is Young's modulus. 

9*3*2 Observed brittle strength: need for cracks 

The observed brittle strength of solids is generally very variable but is 
always 10 to 100 times smaller than the theoretical value. The source of 
this weakness is the presence of flaws or cracks in the solid, especially at the 
surface. These act as 'stress-raisers' or 'stress-multipliers'. Consider a rect- 
angular strip of uniform thickness subjected to a tensile stress a. If there is a 
crack in one edge, as shown in figure 60(6), and we can specify it by its length 
/ and its tip radius p the tensile stress at the tip of the crack is multiplied by a 
factor of the order 

V P 


Hence if the applied stress is a and the theoretical breaking stress is S, we may 



If the crack radius is, say, 10 A. a crack length of 10,000 A. will increase the 
tensile stress at the tip of the crack by 30-fold. Thus an applied stress only one- 
thirtieth of the theoretical strength will be able to start the crack growing. 
Once this occurs the whole section will fail. 

There are three general observations which support this explanation. 
First brittle strengths are usually very variable; this is because of the large 
variety of sizes and shapes of surface cracks. Secondly, in recent years it has 
proved possible to prepare fine silica rods with very perfect surfaces which can 
withstand a tensile stress of £/20 before they fail. Thirdly, some fine crystal- 
line fibres have been prepared which have tensile strengths approaching their 
theoretical value, presumably because their surfaces are free of surface flaws 
or growth steps. 

Although this explanation is generally satisfactory there is an additional 
failure criterion which must be mentioned. Consider a flaw which is in the 

167 The strength properties of solids 

form of a sharp crack between two neighbouring atomic planes; the crack- 
tip radius tends to zero. Is then the stress-concentration factor infinite, and 
does the material now possess zero strength? The answer was given by 
A. A. Griffith in 1921. He showed that for a crack to grow it is not sufficient 
for the stresses at the crack-tip to exceed the theoretical strength; in addition 
sufficient elastic energy must be released from the system to provide the extra 
surface energy that a growing crack demands. By expressing the surface 
energy in terms of the elastic constants, the force-separation curve, and the 
atomic spacing it is possible to estimate what this involves. It turns out [see 
A. H. Cottrell (1964), The Mechanical Properties of Matter, Wiley, London, 
and E. Orowan (1949), 'Fracture and strength of solids', Reports on Progress 
in Physics, vol. 12, 1949, pp. 185-232], that an infinitely sharp crack cannot 
grow with the application of a vanishingly small tensile stress. In order to 
satisfy the surface energy criterion, the smallest stress a, capable of producing 
crack propagation, if the lattice spacing is a, is of order 



that is to say, because of the surface energy criterion an infinitely sharp 
crack produces a reduction in strength equivalent to a tip radius of about 
three atomic spacings. For a crack of length /, the Griffith stress a t is the 
smallest stress that will start the crack growing. This is the stress for a crack 
of tip radius 3a or less. For a more blunt crack (p > 3a) a larger stress is 
needed to start the crack moving. If the geometry of the tip is retained during 
propagation the failure stress falls slowly as the crack length increases. If, 
however, the crack grows into a sharp crack the failure stress falls to the 
Griffith value. 

Equation (9.19) may also be expressed in terms of Young's modulus E and 
the surface energy y of the solid in the following way. The work done in 
pulling an ideally brittle solid specimen of unit cross-section apart is equi- 
valent to the work done in creating two new unit surfaces, i.e. 2 y. As we have 
seen, this pulling apart occurs when the applied stress is of order £/10 [see 
equation (9.16)]. The separating atomic planes are pulled apart at this 

E a Ea 

stage by a distance of the order a/10. The work done is of order = — 

1010 100 

which can be equated to 2y. Hence y is of order Ea/200. Equation (9.19) then 


_£ // 3x200y \ ^ //600 x y£\ 

A more rigorous analysis shows that the numerical factor under the square 
root sign is nearer unity than 6. The Griffith stress then becomes 



168 Gases, liquids and solids 

The equilibrium spreading of a crack thus provides a means of determining 
the surface energy of a solid. We have already described one example of this 
in the splitting of a mica sheet. Attempts have also been made to apply 
this method to metals, polymers and other solids. Generally such experiments 
give extremely high values of the order of 100,000 erg cm. -2 ; this is because 
plastic deformation occurs at the tip of the crack and the plastic work involved 
completely swamps the much smaller true surface energy. Recently Gilman 
has shown that with some ionic crystals it is possible to 'freeze-out' the 
ductility by carrying out the experiments with the specimens immersed in 
liquid nitrogen. Under these conditions surface energies of the order of a 
few hundred erg cm. -2 are obtained. 

Edge dislocations are too small to act as effective stress-raisers from the 
point of view of crack propagation. However, if the free movement of edge 
dislocations is obstructed by some barrier so that several can pile up in a row 
[see figure 59(c)], an internal crack may be formed; this may be able to 
initiate crack propagation. In this way a ductile solid may become brittle. 



' elastic 



Figure 61 . Stress-strain curve for a brittle solid. Brittle failure occurs for a tensile 
stress S. In the presence of a hydrostatic pressure P this is increased to P+S. The 
associated shear stress may produce plastic yielding before brittle failure occurs. 

9 -3 -3 How brittle solids may be made ductile 

Consider a brittle solid which fails in a brittle manner for a tensile stress S. 
If we apply a hydrostatic pressure the tensile stress necessary for failure is 
P+S. Associated with this tensile stress is a shear stress equal to H.P+S). 
If the critical shear stress of the material is less than this it will flow in a 
ductile manner before the tensile stress is large enough to produce brittle 
failure. This is one of the reasons why rocks below the earth can flow in a 
ductile way although they are normally very brittle materials. Indeed Bridgman 

169 The strength properties of solids 

has shown in the laboratory that under sufficiently high hydrostatic pressure 
even quartz can flow plastically. In this connexion it is interesting to note 
that in indentation hardness experiments plastic indentation can often be 
made in relatively brittle materials (even though some cracking may also 
occur). This is because the large hydrostatic component of the stress field 
inhibits brittle failure. Furthermore the hardness values so obtained are a 
measure of the plastic properties of the brittle solid. 

9 -4 Conclusion 

In this chapter we have discussed the strength and deformation properties 
of solids in terms of interatomic forces. The deformation characteristics fall 
into three main categories. When the stresses are below a certain level 
the strains are reversible and the deformation is said to be elastic. As we saw 
in the previous chapter elastic deformation involves the gentle distortion 
of the lattice and when calculations are possible there is good agreement 
between the observed behaviour and that derived from the interatomic forces. 
For larger stresses two other types of deformation occur, plastic and brittle: 
both are irreversible. These strength properties may also be calculated in 
terms of interatomic forces but observed values are generally very much 
smaller. We may say that solids owe their plastic and brittle strength to 
interatomic forces: their weakness to the presence of flaws or imperfections. 
The main characteristics are summarized in the following table. 

Table 22 Plastic and brittle failure 

Mode Criterion 

Source of weakness 

Effect of hydro- 
static pressure 

Plastic Critical shear stress 
for planes to slide 
over one another 

Dislocations, en- 
able slip to occur 
row by row of 


Brittle Tensile stress across Cracks, act as 

a plane pulling 
planes of atoms 

so that stress at 
crack tip is much 
greater than the 
applied stress 

Opposes applied 
tensile stress and 
so can inhibit 
brittle failure 

170 Gases, liquids and solids 

Chapter 10 

Thermal and electrical 

properties of solids 

In this chapter we shall discuss in terms of atomic mechanisms the specific 
heat of solids, thermal expansion and thermal and electrical conductivity. 

10-1 Specific heat 

10-1-1 Definition of specific heat 

The specific heat or heat capacity is defined as the quantity of heat required to 
raise the temperature of one gram of substance by 1°C. With solids the specific 
heat at constant volume C v is a little less than that at constant pressure C P . 
For a gram-molecule the corresponding quantities are related by the thermo- 
dynamic function 

B 1 TV 

Cr -Cy = ^~, (10.1) 

where fi is the volume coefficient of expansion, K the bulk modulus and V the 
volume of one gram-molecule. In practice the specific heat generally measured 
with solids is C r , but the difference between this and C r is small and can as a 
first approximation be neglected. 

If we assume that the specific heat is due to the change in internal energy 
we may easily calculate it, say, for a gram-atom of a metal. This contains JV* 
atoms. The atoms have no translational energy, only vibrational. Each vibra- 
tional degree of freedom involves both potential and kinetic energy so that the 
average thermal energy is kT per degree of vibrational freedom. Each atom 
has three independent degrees of vibration so that its average vibrational 
energy is 3kT. For all N atoms this gives a value of 

U = JVo x 3kT = 3RT. (10.2) 

If we keep the volume constant any additional heat goes solely in increasing 
the vibrational energy. Hence 


3R, (10.3) 

where C r is the specific heat per gram-atom. Thus C v should have a value of 
about 6 cal. per degree, a conclusion reached empirically by Dulong and Petit 
in 1819. If one deals with a compound, say NaCl, then one would expect each 
gram-ion to possess a specific heat of 6 cal. °C. -1 ; so that for the mole 

171 Thermal and electrical properties of solids 

the specific heat would be 12 cal. °C. - * ; for a triatomic molecule it would be 
18 cal. "C. -1 . Some typical results for C, at room temperature for metals and 
diatomic and triatomic solids are given in Table 23. Ignoring the difference 
between C, and C v , it is seen that the specific heats per gram-atom are all 
indeed of the order of 6 cal. °C. -1 (except ice). 

Table 23 Specific heats of solids 


cal. gr 1 °C.-» 

Atomic or 
Molecular weight 

Specific heat C P 
g-atonr 1 or mole' 1 
ad.°Cr l 







CaF 2 
SiO a 



12-0 for 2 gram-atom 

17-1 for 3 gram-atom 

16-8 for 3 gram-atom 

9-0 for 3 gram-atom 

10-1-2 The Einstein model 

Later measurements showed that the Dulong and Petit law holds only above 
a certain temperature. At lower temperatures C, falls off markedly. For 
example with copper the specific heat per gram-atom falls off from 5-7 at room 
temperature to 51 at -100°C., to 2-7 at -200°C. and to 0-2 at -250°C. 
In the limit it approaches zero at the absolute zero. 

The first simple explanation of this was due to Einstein in 1906 who made 
use of Planck's earlier discovery of the distribution of thermal energy in an 
oscillating system. If each atom behaves as an independent harmonic oscil- 
lator of frequency v, its average energy (ignoring zero-point energy) is exactly 
the same as that given in our discussion of the vibrational energy of a di- 
atomic molecule. 

u = 





For the N atoms each of which has 3 independent degrees of vibrational 

172 Gases, liquids and solids 

freedom we obtain 
U = 3N u. 


For high temperatures this reduces to 3N kT in agreement with the Dulong 
and Petit law. For lower temperatures it diminishes and as T tends to zero, 
it also tends to zero. If an appropriate value of v is assumed the shape of the 
Einstein theoretical curve agrees reasonably well with experiment except 
towards T = (see figure 62). In this region careful experiments show that 


t a ■ 




















temperature °K. 

• experimental 



theory (Einstein) assuming fV=240°K 

Figure 62. Specific heat of copper as a function of temperature : • experimental 
values, -theory given by Einstein assuming an atomic vibrational frequency of 

v = 5x 10' 2 sec. -1 If one writes ^ ■■ 

k0 E 
corresponding to v = 5x10 12 sec."', a value of 9 = 240°K. 

1 this gives for the Einstein temperature. 

C r falls off as T 3 , whereas in Einstein's theory it falls off as exp f J , i.e., 

it falls off more rapidly. 

The Debye model 

There are two main defects in the Einstein model. First it assumes that all the 
atoms have the same single frequency v, and secondly that the vibration of 
each atom is independent of its neighbour. In fact the atoms act as coupled 
oscillators and a whole range of frequencies is possible. Each value of v (at a 
fixed temperature) contributes its own average thermal energy « [see equation 
(10.4)]. These should all be added to determine the total energy J/of the whole 
solid. An analysis along these lines was first developed by Born and von 

173 Thermal and electrical properties of solids 

Karman in 1912. The main feature of such a treatment, which distinguishes 
it from the Einstein model, is that it reveals a whole spectrum of possible 
frequencies. The detailed analysis is, however, extremely complicated. 

A completely different approach is that due to Debye (1912). The solid is 
treated as a continuum which, at first sight, seems a retrogressive step. 
However, its great merit is that it provides a frequency spectrum which is a 
close approximation to the true spectrum. Debye considers the way in which 
waves can travel through the solid. The most remarkable feature is that a 
standing wave of frequency v behaves exactly like a quantum oscillator of fre- 
quency v so that its thermal energy is again given by equation (10.4). This is 
surprising since the wave involves the vibration of all the atoms in the solid; 
yet its thermal energy is that of a single quantum oscillator. At first sight it 
would seem that this could not possibly give the same total energy. The diffi- 
culty, however, disappears when it is realized that the total vibration of each 
atom is the composite result of all the possible waves that can travel through 
the solid. 

Suppose the solid is a crystal in the form of a cube of side L. To establish 
standing waves the free surface of the specimen must be either a node or an 
antinode and it may be shown that either condition leads to the same result. 
We shall treat the free surface as an antinode. Then the largest standing wave 
possible for a wave travelling normal to the faces of the cube has a wavelength 
A = 2L (see figure 63). If the velocity of the wave is C the frequency of the 

C C 

wave is v = — = — . The next possible standing wave has a wavelength 


£ -£. 1 

2i 2i 

" L 2L Z 

3 C C , 

2 r = 2i a 

Figure 63. Characteristic frequencies of standing waves in an elastic continuum 
of length L. We assume that the free ends are antinodes and that the velocity C 
of the waves is independent of frequency. 

174 Gases, liquids and solids 

A = L and its frequency is v = C/L. The next has a wavelength A = \L and its 

3 C 
frequency is v = - x — . Thus the possible frequencies increase in the order: 
2 L* 

C C C C 

— ; — x2; — x3; — x 4, etc. or vj, 2v u 3vi, 4vi. (10.6) 

2L 2L 2L 22, 

The corresponding thermal energies are 
hvi 2hv t 


exp^-j-1 exp^—j-1 exp^—j- 

,etc. (10.7) 


and all that is now needed is to sum them. Very soon the possible frequencies 
become such large multiples of v t that they become virtually continuous and 
the sum can be replaced by an integral. One needs to know the number of 
vibrations that are possible between v and v+dv. This is not difficult for a 
continuum. A simple approach is as follows. 
For the waves discussed above 


" 2L 

where « is an integer. If we considered the more general case of a wave tra- 
velling in some arbitrary direction we would obtain the same result but n 
would have components n lt n 2 , n 3 (each of them integers) in the x, y, z 



a» — •-- JL •"'**•'».• • • • • 

o • • •"•*.• ^ • • • 

N. s 

!>•••• •"„• -• • • 


/\ ^ 

4) • • • •0^» «\ • \ • 

JTV \ \ 

it e • •/* • • • f f 


Figure 64. (a) Construction showing that the number of waves dN possessing 
frequency between v and v+dv is proportional to the volume in the octet of the 
spherical shell lying between radii v and v+dv. (/») Two-dimensional figure show- 
ing that each point corresponding to a particula r value of v occupies a cube of 
side (C/2L). 

175 Thermal and electrical properties of solids 


v = £(«?+«!+»!)*• (10.8) 

c c c 

We plot «i — , «2 — . «3 — on x, y, z co-ordinates. Then the distance from 
4JL1 ZXi iJ-t 

the origin to any point is equal to v. Since n u n 2 , n 3 can only change by one 

unit at a time each possible point for v occupies a volume of j — J . The 

vailable for frequencies between v and v+ 
ell contained between radii v, v+dv divi 

the spheri 

dJT = (ix4m 2 dv) PgY 

= ~Z • (10.9) 

We allow for two transverse vibrations and one longitudinal vibration. If 
both have the same wave velocity C (or if we choose a suitable average 
velocity) this triples the value of djf. Then the total thermal energy is simply 

number of points available for frequencies between v and v + dv is the volume 
of the spherical shell contained between radii v, v+dv divided by the unit 

volume j — J . Since v can have only positive values we can only use one- 
eighth of the spherical shell. The resulting number is therefore 




i<ur. (10.10) 

We have ignored zero-point energy since it is independent of rand when we 
differentiate U to find C v it will disappear. 

The difficulty about equation (10.10) is that we have no way, so far, of 
specifying the upper limit v„ of the integral. It is at this point that Debye ties 
his continuum model to a particulate model. If the crystal contains N atoms, 
Debye postulates that the total number of vibrations possible must be equal 
to 3iV~ so that, in the limit, at high temperatures where each u has a value of 
kT the total thermal energy is 3NkT. From equation (105) this means that 

f TxUf = f" 
Jo Jo 

\2n^- 3 v 2 dv = 3iV. (10.11) 

This gives 

v ~=h$ N - (10 - 12) 

Before we use this we may form some estimate of its magnitude. If the solid 
176 Gases, liquids and solids 

has a simple cubic structure with a distance a between each atom, Na 3 will be 
the volume of the solid, i.e., L 3 . Then 





This corresponds to a minimum wavelength 

^•mln — 


Thus the limiting frequency on the Debye model corresponds to very short 
waves, the wavelength being comparable with the atomic spacing. This is 
reasonable since once we recognize the particulate nature of the solid, a wave- 
length less than the atomic spacing ceases to be meaningful. This result also 
corresponds rather closely to the very simple elastic model discussed in the 
previous chapter. We found that if the atoms have a natural (uncoupled) fre- 
quency v the velocity of a wave is of order 2nva compared with v m a given in 
equation (10.13). Inserting equation (10.12) in (10.9), 



If this is substituted in equation (10.10) we have an explicit expression for U 

which can be evaluated. By differentiating, the specific heat C v may be found. 

The Debye theory predicts that the heat capacity of a solid depends only 


O Cu:0o=315°K. 
• Ag:0D=215°K. 

■ Pb:0 D =88°K. 
x C :0 D =186OK. 


Figure 65. Specific heat of various solids plotted as a function of 7— where v m 

nv m 

is the characteristic Debye frequency. Writing —t- = 6 D , the quantity 6 D is known 

as the Oebye temperature. 

Q • ■ x experimental points for Cu, Ag, Pb, diamond theory given by 


177 Thermal and electrical properties of solids 

on the characteristic frequency v m . This implies that if C r for various solids 

is plotted against — they should all lie on a single curve. This is found to be 

very nearly true (see figure 65). The quantity — is known as the Debye tem- 


perature 6 D . It is interesting to compare B D as calculated from equation 
(10.12) with the value of D that gives the best fit of the Debye equation 
with the observed specific heat measurements. 

Table 24 The Debye temperature 




From specific 

From equation 


















CaF 2 






Although the theory is only really valid for isotropic solids containing a 
single type of atom it is seen that it holds extremely well for more complicated 
materials. It even holds approximately for ice, which is rather surprising. 
(An average wave velocity of 2-2 x 10 5 cm. sec. -1 has been used to calculate 
D ).The specific heat of ice at 200°K is about 0-35 cal. g. - 1 or 6 cal. mole - x . 
It thus behaves as a monatomic solid, practically the whole of the thermal 
energy being due to oscillations of the molecule as a whole. Between 200°K 
and 273°K (the melting point) there is some increase in specific heat to a 
value of about 9 cal. mole -1 . Presumably in this temperature range rotat- 
ional degrees of freedom of the whole molecule contribute to the thermal 
energy of the ice crystal. As we shall see in Chapter 13 this rotational 
freedom accounts for the rather large low-frequency dielectric constant of 
well ice below 0°C (see figure 97). 

Although the Einstein model is in many ways a simpler and more direct 
one than the Debye model there are two features in the Debye treatment that 
make it preferable. First v m may be derived directly from bulk properties so 
that there are no assumed constants in the final calculation of V. By contrast 
the Einstein frequency (which turns out to be comparable with Debye's v„, 

178 Gases, liquids and solids 

in fact for many solids, v E s 0-75v M or 6 E 2; O-750 D ) must be deduced empiri- 
cally from the shape of the Cy~T curve. Secondly at low temperatures the 
Debye relation for the specific heat reduces to 

12»i 4 i? „. 


so that it satisfactorily explains the observed T 3 dependence in this temperature 

The main defect of the Debye treatment is that it considers waves of all 
frequencies to travel at the same speed. This is not true of coupled oscillators. 
For higher frequencies there is an appreciable drop in wave velocity. For- 
tunately the waves in this range generally contribute only a small part to the 
total vibrational energy. 

10'2 Thermal expansion: Gruneisen's law 

We may now derive a simple relation between thermal expansion and specific 
heat. It is based on the existence of an asymmetrical potential energy curve 
for the whole crystal in terms of the separation between atoms. For simplicity 
we consider here only the potential energy u between one atom and its neigh- 
bour and ignore the problem of coupled interactions. 
The potential energy curve is shown in figure 66(a) ; we transpose it to axes 

Figure 66. (a) Potential energy u of an atom as a function of its separation x from 
its neighbours, (b) same curve transposed to different axes so that the minimum 
has coordinates 0, 0. 

179 Thermal and electrical properties of solids 

Figure 66. (c) force-separation curve for the situation described in (6) ; the expan- 
sion o is equivalent to an increase dx in the mean spacing. 

as shown in figure 66(b) and express u in terms of the displacement x of the 
atoms from their equilibrium separation a. We assume that at 0°K. the energy 
is zero (ignoring zero-point energy) and that the addition of thermal energy 
increases u according to a power law: 

u = Ax+Bx 2 +Cx 3 .... 


Since O is the origin where — = we obtain A = 0. Then 


u = Bx 2 + Cx 3 + 



If we used only u = Bx 2 the curve would be symmetrical about the u axis and 

the restoring force for small displacements, = — 2Bx, would be pro- 
portional to x, so that the atoms would oscillate with simple harmonic motion. 
We now assume that the cubic term is sufficient to match the real asymmetric 
potential energy curve. 

u = Bx 2 + Cx*. 


At a value of energy u t there are two possible values of x; we call them 
Xi and — x 2 . 

u t = Bxf+Cxl = Bxl-Cxl 
Bixf-x 2 ) = -Cr>f+*D. 


180 Gases, liquids and solids 

When adding x t and x z they can be considered as being almost equal. 
Equation (10.20) becomes 

B(x l -x 2 ) = -Cx 2 . (10.21) 

Oscillations between P and Q are not strictly simple harmonic because of the 
cubic term in x. The mean position, however, is approximately the mid-point 
of PQ. Its distance e from the u axis is K*i — ^2). 
Hence, from (10.21) 

The quantity e is the thermal expansion of the lattice when the mean vibra- 
tional amplitude is x. Its variation with temperature is 

de -C d , , s 

Now the thermal energy of vibration in one dimension is approximately 

u = Bx 1 so that the specific heat for a one-dimensional vibration is 

8u d 

— = B — (x 1 ). Since this involves only the vibrational energy, it is the speci- 
fy at 

fie heat for constant volume. This makes no assumptions about equipartition 

of thermal energy so that for this model it holds however far the material 

may be from the Debye temperature. The specific heat per atom in three 

dimensions, c v , is three times this, so that 

<V- IB j t {x 2 ). (10.24) 

Hence from (10.23) 

de -C 

dJ = W Cy ' < 1025 > 

The coefficient of linear expansion a is given by 

1 de -CI 
a = adf = W X a Cy - <***> 

Since a varies little with temperature we see that a is directly proportional to 
the specific heat at that temperature. The greater the specific heat, the greater 
the thermal vibration, and the greater the effect of the cubic term in augment- 
ing the expansion. This is the main feature of the Gruneisen treatment. 

It is possible to express B and C in terms of other bulk properties of the 
solid. For example, for a simple cubic array B is equal to Ea/2, where E is 
Young's modulus. The other parameter chosen by Gruneisen is the rate of 
change of vibrational frequency with lattice expansion. The natural frequency 

181 Thermal and electrical properties of solids 

of vibration v is proportional to (force constant)*, i.e. to ( — J Hence 


log v = constant +i log — - . 
8x 2 

d .. . 1 / 1 \ 8 3 u 
^^■iKp- (10 ' 27) 

From equation (10.18) we have 

8 2 u 

— m 2B+6Cx ~ 25, J 

dx 2 dx- 


8 2 u 8 3 u 

— - 2B+6Cx ~ IB; — - % = 6C. (10.28) 

</ , , . 1 6C 3C 

— log (v) = - X = . 

dx sw 2 25 25 

But — (log«) = -, (10.29) 

dx a 

since on the force-displacement curve [figure 66(c)] the displacement of the 
equilibrium distance dx is the change in the spacing. 

rflogv 3Ca # .„-«v 

—~- = —Z7T- ( 10 - 3 °) 

d log a 25 

The Gruneisen constant y is usually defined in terms of the change of fre- 
quency with atomic volume v. Since v is proportional to a 3 we have 

rflogv rflogv = _C* 

7 dlog* 3d log a 2B' {W31) 

Substituting from equation (10.31) for C — and B = — in equation 

a 2 

(10.26) we obtain 

«-3§£i«r. O0.32) 

For a simple cubic structure Nod 3 is the volume K of a gram-atom and N c v 
is the specific heat Cy of a gram-atom. Hence 

2y C v 
" = 3£ X 7 (1<U3) 

182 Gases, liquids and solids 

at x = a, 


(The coefficient of cubical expansion S is 3a so that 
. 2yC r 

A more rigorous model gives 

where K is the bulk modulus.) 

We see that a is expressed solely in terms of measurable bulk quantities. 
As a matter of interest we may calculate a for an ionic solid for which the 
potential energy between a pair of ions is 

«=-*[--^V|. (10-34) 

""&?" k [x> *»J" a* 

- 8 3 u , r-6 110fl 8 1 , 104 

From equations (10.27) and (10.29) we see that 

y = -1JW1 = _ *!?11 = _?". (10 .35) 

d log v 3d log a 6 « 

Consequently for an ionic solid 

a 104a 3 _ 13 
y= ~6 X 8a* 6* 

For rock-salt E s: 2-10 11 dyne cm. -2 ; V 2: 40 cm 3 mole -1 ; 

C v x C, =s 50-10 7 erg "C." 1 mole- 1 . 

From (1033) this gives 

a 2 10-* "C.- 1 

The experimental value is of this order but smaller by a factor of about 4. 
With non-ionic solids y is~appreciably less and a is about 10 times smaller. 
As it has been pointed out at the end of Chapter 7, the above treatment 
has one basic defect; namely that it treats the solid as though its behaviour 
were exactly analogous to that of a pair of atoms. By contrast when we consider 
the behaviour of an assembly of a large number of atoms as in a solid crystal, 
an entropy factor arises which is not accounted for by the simple treatment. 
It may indeed be shown that even if the atomic vibrations were perfectly 
harmonic, that is if the potential energy curve were perfectly symmetrical, 

183 Thermal and electrical properties of solids 

the increase in free energy of the crystal as its temperature is raised would 
lead to an increase in volume [E. A. Guggenheim (1965), Boltzmann's Dis- 
tribution Law, North-Holland Publishing Company]. However the thermal 
expansion due to this factor appears to be small compared with that arising 
from the asymmetric nature of the potential energy curve considered above. 

10-3 Thermal conductivity 

10-3-1 Heat transfer 

There are three main forms of heat transfer. 

(a) Radiation; this involves the emission of electromagnetic waves. All 
materials can radiate but radiation is generally most marked with solids. 

(b) Convection; this involves the streaming of matter. With forced con- 
vection the streaming is imposed by some external driving mechanism; 
with natural convection it occurs as a result of density differences. Con- 
vection is observed only in liquids and gases. 

(c) Conduction; this involves the transfer of thermal energy by some form 
of collision. It occurs in gases, liquids and solids. With poorly conduct- 
ing solids the energy transfer is provided by lattice vibrations; with 
metals this is greatly augmented by electron collisions. In this chapter 
we shall deal solely with conduction in solids. 

The rate at which heat flows through any element of a solid is proportional 
to the cross-sectional area dA of the element and the temperature gradient 
measured normal to dA. We may make this proportionality into an equality 
by inserting a constant K which we call the thermal conductivity. Then 

-f = -KdAx — . (10.36) 

dt dx 

The negative sign shows that a positive flow of heat takes place from a hotter 
to a cooler region, i.e., where dT/dx is negative. K has the dimensions 
MLT~ 3 Q- 1 . It varies from the best thermal insulators to the best conductors 
by a factor of only about 1000. By contrast, electrical conductivities can vary 
by a factor of 10 30 . 

10-3-2 Heat flow down a uniform bar 

As a simple example we consider the heat flow down a uniform bar under two 
extreme conditions. When it is 

(a) thermally insulated. In the steady state if no heat escapes from the 
surface of the bar the quantity of heat crossing every section per second 
must be constant. The temperature gradient must therefore be constant 

184 Gases, liquids and solids 

[see equation (10.36)]. Knowing the temperature of the hot and cold 
ends, the cross-section of the bar and the rate of heat conduction, K 
may be determined directly. This is a convenient method of measuring 
K for good conductors. It may also be used with poor conductors if the 
bar is replaced by a thin disc of the material. 

(a) - 


A! I !B 



Figure 67. (a) Steady-state flow of heat along an exposed uniform bar. The heat 
flow in at A, less the heat flow out at B is equal to the heat lost by radiation and 
convection from the surface between A and B. (b) Heat flow through a thermally 
insulated element : the heat flow in at A, less the heat flow out at B is equal to the 
heat accumulated in the element 

(b) exposed to a cooler constant temperature environment. Consider a 
very long bar of uniform cross-sectional area A held at a temperature 
7" at one end (figure 67). Let the environment be at a temperature T e . 
Consider a small element of length dx at some point along the bar where 
its temperature is T. Express these as temperature excesses over the 
environmental temperature. The hot end is at 6 , the element at and 
the environment at 0°. In the steady state heat enters at A, some is lost 
by radiation and convection from the surface of the element, and the 
remainder passes through the element at B. If the circumference of the 

185 Thermal and electrical properties of solids 

bar is c the surface area of the element is cdx. If the heat lost is pro- 
portional to the surface area, the emissivity a of the surface, and the 
temperature excess over the surroundings, the heat lost per sec. is 
cdx eO. 


Heat flow per sec. in at A = —KA — . 


Heat flow per sec. out at B = -KA^iO+dO) = -KAl—+—,dx\. 

dx \dx dx 2 ) 

Heat flow per sec. at A exceeds outflow at B by 

d 2 
KA — - dx = heat lost from surface of element. 
ox 2 

KA d -^ = ced. (10.37) 

dx 2 

The solution is of the form 


where w 2 = — . At great distances from the hot end the temperature ap- 

proaches the ambient, i.e., = 0. Hence B = 

d = Co- 1 **. (10.38) 

Since at x = 0, = 9 we have finally 

= o exp -hgj) x . (10.39) 

The temperature falls off exponentially with distance from the hot end. 

10-3-3 Thermal diffusivity 

Consider a body which is insulated so that no heat is lost to the surroundings. 
We heat one part of it and consider how the heat flows before steady-state 
conditions are reached. Clearly the body has to be warmed up (i.e., it must 
absorb heat) as the heat flows through it. 

Consider a small element of cross-section dA at a point A (ordinate x) 
where the temperature is T. At the neighbouring point B (ordinate x+dx) 
the temperature is T+dT [see figure 67(b)]. 

Heat flow per sec. in at A = — KdA — . 


Heat flow per sec. out at B = -KdA — (T+dT). 


186 Gases, liquids and solids 

The heat accumulated in the element per sec. = KdA — - dx. If the element 

has density p, specific heat C, its heat capacity the temperature 

rise per sec. of the element is — we have 


dT 8 2 T 

p CdA dx^r = KdA — dx, (10.40) 

dt dx 1 

dT = K 8*T 

dt pC dx 2 ' 

The quantity — is known as the thermal diffusivity or temperature conduc- 

tivity. It is of great importance in non-steady-state thermal problems. 

10-3 -4 Theory of thermal conductivity 

The thermal conductivity of solids can be explained in terms of some type 
of collision process by means of which thermal energy is transported from 
the hotter to the colder regions. It is natural therefore to express it in a 
manner resembling the treatment for gases. For a gas where the transfer is 
due to molecular collisions we have 

K = inmcXCr, (10.41) 

where C v is the specific heat per unit mass. We note that the quantity nmCy 
is the specific heat per unit volume s. Then 

K = iscX. (10.42) 

0-3-4-1 Poor conductors. In poor conductors the thermal energy is transferred by 
lattice vibrations. These travel through the specimen as waves which are scat- 
tered by the lattice as they progress through it. The wave is attenuated corre- 
sponding to the drop in temperature through the specimen. These thermal 
waves thus behave like particles possessing energy and momentum: they are 
known as phonons. 

We may treat the waves as energy packets moving with the velocity c of a 
sound wave, and travelling a distance A before they collide and communicate 
their thermal energy to the lattice. By exact analogy with gases we may write 

K = ^{specific heat per unit volume) (velocity of sound wave)A. 

This relation is found to be quite satisfactory for most poor conductors, the 
value of A being of the order of 20 A. 

D'3-4'2 Good thermal conductors: metals. Good thermal conductors are generally good 
electrical conductors - their thermal conductivity is 100 to 1000 times greater 

187 Thermal and electrical properties of solids 

than that of insulators. This suggests that the major part of the thermal 
conductivity, like that of the electrical conductivity, arises from movement 
of electrons. 
We make the following assumptions about the metallic state: 

(i) The metal consists of fixed positive ions in a sea of electrons : in general 
there will be one or two electrons per ion. 

(ii) The electrons behave as a perfect gas transporting thermal energy as 
in gas conduction. 

(iii) Each electron has thermal energy of translation equal to \kT, i.e., its 
thermal capacity is %k. 

(iv) Each electron travels a distance X before colliding with a positive ion 
and giving up ail its thermal energy. 

We rewrite equation (10.41) in the form 

K = incX(mC r ), (10.42a) 

and note that mC v is the thermal capacity of a single travelling particle. 
Replacing this by \ k for our electron gas, we obtain 

K = incXk, (10.43) 

where c is the velocity of an electron assuming it to behave as a gas molecule. 
At room temperature its value is of order 10' cm. sec." 1 If we consider a 
monovalent metal such as sodium, for which there is a single electron per ion, 
we have n sl 50 x 10", k = l-4x 10" ". The observed conductivity at room 
temperature is about l-3x 10 7 erg cm. °C. sec. -1 This gives A 2: 3 A. In 
general, equation (10.43) gives a good agreement with observation on the 
assumption that X is of the order of the atomic spacing. Although this seems 
satisfactory there is a major defect in the theory which we shall discuss later. 

10-4 Electrical conductivity of metals 

10-4- 1 Drude model 

The general picture that we adopt here, which is due to Drude (1912), is 
similar to that used in the preceding section. It is assumed that the electrons 
move like a perfect gas in all directions through the lattice and that their 
thermal motion is terminated by collisions with positive ions. Under normal 
conditions there is no net transfer of charge in any direction. When, however, 
a potential is applied across the metal the electric field imposes a drift velocity 
on the electrons and it is this which is responsible for the observed electrical 
conductance. The drift velocity is very small compared with the mean electron- 
gas velocity c. 
Consider a bar of uniform cross-section A, length L, across which a potential 

188 Gases, liquids and solids 



Figure 68. Electrical conductivity model for metallic conductors. 

V is applied [figure 68(6)]. The electric field on the electrons is X = —.The 



force on an electron is eX and its acceleration is a = — . If X is the mean free 


path of the electron the time between collisions is t = X/c. The drift velocity 

acquired during this period is ar so that the average drift velocity u between 

collisions is ±at. 

« = $«, = ---. 
2 m c 


The model still assumes that the electron gives up the velocity gained from 
the field by colliding with a positive ion. If #i is the number of free electrons 
per c.c. the number travelling per sec. along the bar due to drift is the number 
in a volume Au, i.e., Aim. Each electron carries a charge « so that the charge 
carried per sec., i.e., the current /, is 

/ = eAun 

_ 1 AXle 2 n 
2 mc 



, AV Xe 2 n 
L 2ml 


If p is the specific resistivity and * the specific conductivity we know that 


AV 1 AV 

i = — x - or — x k . 

L p L 

Comparison with (10.46) shows that 

- *£f? - ^^ 
2mc 2md 2 ' 


189 Thermal and electrical properties of solids 

If we ignore the difference between «* and H* we may, for the electron gas, 

imc 2 = IkT. dO.49) 

Hence, from equation (10.48) 

Ae 2 nd 

For a monovalent metal assuming A = 3 A., n = , e ~ 1 -6 x 10 _2 ° 

27xl0 -2 * 

e.m.u., c = 10 7 cm. sec. -1 , k = l-4x 10 -16 , T = 300°K., 

k m 10" 5 e.m.u. (10.51) 

This is in reasonable agreement with the resistivity values of most metals, 
p as (2 to 20) x 10" 6 ohm cm., or conductivity = 5 x 10" * to 5 x 10" 5 e.m.u. 

10-4-2 Wiedemann-Franz relation 

We may combine equations (10.43) and (10.50) to find the ratio of thermal to 
electrical conductivity. We have 



T. (1032) 

Historically, Wiedemann and Franz observed that K/k is the same for all 
metals at the same temperature. This was extended by Lorenz who observed 
that the ratio was proportional to the absolute temperature. The above 
analysis contains both 'laws*. The theoretical ratio is indeed close to the 
observed values over an appreciable temperature range for a large number of 

10-4-3 Specific heat of the free electron 

In the Drude model discussed above the electron gas is assumed to possess 
thermal energy. Like any monatomic gas it should therefore contribute a 
specific heat of \ R. The specific heat of a monovalent metal should therefore 
consist of 3/J from its vibrational (lattice) energy and $R from its free electrons, 
i.e., the specific heat should have a value of about 9 cal. g-atom -1 °C. -1 The 
observed value in fact does not exceed 6. 

This is a basic defect of the electron gas model and the explanation can only 
be given in terms of the energy states of the electrons in a metal. No energy 
state can contain more than two electrons (of opposite spins), so that the 
electrons fill the available states from the lowest available energy level until 
they reach the topmost or Fermi level, £> (see figure 69). This is the situation 

190 Gases, liquids and solids 

at absolute zero. If the metal is heated only those electrons, which can be 
moved into a higher vacant state above E F , can acquire further energy. In 
effect this means that only the electrons at the top of the Fermi level can 
acquire energy. Analysis shows that if there are n free electrons per c.c. the 
number of thermally excited electrons at temperature T is 

An ~ «x—- . 


For a typical metal E p si 2 eV. and at room temperature (T a: 300°K.) Wis 
about 4 xlO -14 erg ~ 0-025 eV. Thus Aw is only about 1 per cent of «. If the 
thermal energy acquired by these electrons is of order A:Twe see that the specific 
heat is only about 1 per cent of the value quoted above. It is for this reason that 
the electronic contribution to the specific heat is so small. 




Figure 69. Electronic energy levels in a metal shown as -~ plotted against the 

energy E so that the shaded area represents the number of electrons per c.c. with 
energy between E and E+dE. (a) At absolute zero all the energy states are filled up 
to a level Ep known as the Fermi level, (b) At a higher temperature T a small number 
of electrons at the top of the Fermi level are able to acquire additional energy of 
order kT. 

We may take this a stage further. Since the specific heat of a single electron 
is of the order k, we see from equation (10.53) that the specific heat per c.c. 
is of order 

*An = 

nk 2 T 


so we may regard the average specific heat per electron as 
k 2 T 


191 Thermal and electrical properties of solids 

The expression from equation (10.42a) for the thermal conductivity may be 

K = ineMcJ, (10.56) 

whibt the electrical conductivity, from equation (10.48), may be written 

* = 2^- (10.57) 

The ratio is 

K 2mf J , , 

= ^-T<0- (1058) 

k 3 


(At this point if we assume the electrons behave like a perfect gas mc 2 is 2kT 
and c, = \k\ we then obtain precisely the classical Drude result given in 
equation (1052).) 

The quantity c, the velocity of those electrons in the Fermi distribution 
which take part in the thermal transport process, is of order 

*2 E F 

s -„* (W59) 

so that equation (1058) becomes, after substituting for c, from equation 

K E F/ . k 2 T 

- " ^ (c -> ~ ^ ' 00.60) 

A more rigorous treatment gives a constant l^- J in front of this expression 

so that it is very nearly the same as equation (1052) obtained from the simple 
Drude model. 

The conclusions one derives from this, however, expose another issue of 
very great importance. The value of E F of about 2 e V implies that the electrons 
have velocities [see equation (1059)] about 10 times greater than the free 
electron-gas velocity. To explain the observed conductivities one must assume 
a mean free path X of the order of a few 100 A. On the Drude model A has a 
value comparable to the distance between the positive ions in the lattice; this 
seems reasonable. A value of several 100 A. seems strange. The explanation 
lies in the fact that electrons travelling through a perfect lattice experience no 
resistance whatsoever. If we treat the electrons as waves it is possible to show 
that electrons scattered by the lattice recombine to give a wave of undiminished 
amplitude: in other words there is no attenuation of the wave so that the 
resistance is zero. At norma) temperatures the lattice vibrations render the 
lattice irregular so that the electron waves are scattered in an irregular way. 
It is this which accounts for the observed finite resistance. Any factor which 
increases the irregularity of the lattice increases the resistance. This is why 
the resistance increases with temperature and with impurity content. 

192 Gases, liquids and solids 

Chapter 11 
The liquid state 

The main characteristics of the gaseous state are random, translational motion 
of the molecules and relatively little molecular interaction. In the solid state 
the main characteristics are fixed sites for the individual atoms or molecules 
and very little translational movement. Both these states are well understood. 
By contrast the liquid state, in some ways, has 'no right to exist'. The molecular 
interaction must be quite strong since a given quantity of liquid, like a solid 
(and unlike a gas) occupies a definite volume. On the other hand the molecules 
have so much freedom that a liquid, unlike a solid, flows very readily and 
easily takes up the shape of the vessel it occupies. 

The gaseous state has been thoroughly investigated and most of its prob- 
lems have been satisfactorily resolved. The solid state is still being studied in 
hundreds of physics laboratories all over the world. By contrast the liquid 
state is a neglected step-child of the physical scientists. This is partly because 
the liquid state raises a number of very difficult theoretical problems. It is also 
partly a matter of fashion dictated by the economic value and engineering 
uses of solids as structural materials, electrical devices, etc. But for chemists, 
physical chemists and especially biologists, the liquid state is possibly of 
greater importance than the solid state. We may well expect, we should cer- 
tainly hope for, marked advances in our understanding of the liquid state in 
the next two or three decades. 

There are basically three approaches to the liquid state. The first is to treat 
it as a modified gas; the second is to treat it as a modified solid; the third 
is to treat it sui g eneris - this is the most difficult of all. 

11-1 The liquid as a modified gas 

A liquid in static equilibrium cannot withstand a shear stress. If we compress 
a column of liquid with a piston the situation is not like that of a solid where 
a uniaxial pressure or tension produces a shear stress in the medium. Unlike 
a solid the liquid must be held in a container; the walls exert pressures on the 
liquid in such a way that no resultant shear stress remains. This is equivalent 
to saying that the only stress the liquid can experience is a hydrostatic pressure 
or tension. It is for this reason that, at any point in a stationary liquid, the 
pressure is the same in all directions. In this sense a liquid closely resembles a 
gas and indeed both states are sometimes designated by the single generic 
term 'fluid'. In terms of resistance to pressure it therefore seems reasonable 

193 The liquid state 

to consider how far the gas laws can be applied to explain the behaviour of 
liquids. Naturally we would not expect the ideal gas laws to be applicable 
but we might expect the imperfect gas laws to have some relevance. 

We start by considering a liquid as a limiting case of a van der Waals gas. 
We ask what meaning may be attached to the van der Waals constants in 
the equation 

(p+fy(V-b) = RT. 

(a) The quantity b. In a gas b = Av m and at the critical volume V e = 36 so 
that the quantity ( V e — b) is still positive and meaningful. At temperatures 
well below the critical temperature the density of the liquid increases, 
the volume occupied by the liquid is only perhaps (1 -5 to 2)v m so that 
{V— b) is negative unless a different correction value is given to b. 

(b) The quantity a\V 2 . So long as the molecular volume is large a/V 2 is 
small and is meaningful as a correction term. However, when V is 
comparable with or less than b the term a/ V 1 has a value of the order of 
1000-10,000 atmospheres. This internal pressure is so large compared 
with the atmospheric pressure that it cannot be treated as a correction 
factor. This is, of course, consistent with the fact that the external 
pressure P is not an important factor in determining the volume occupied 
by a liquid. 

(c) The tensile strength of a liquid. In what follows we shall show that for 
low-temperature isotherms a liquid, according to van der Waals* 
equation, can actually withstand negative pressures. If we apply an 
increasing negative pressure (hydrostatic tension) the liquid will expand 
until at some critical value vapour will begin to form within the liquid, 
that is to say the liquid begins to cavitate or pull apart. The pressure 
at which this occurs provides, therefore, a measure of the tensile strength 
of the liquid. 

Following Cottrell (1965) we use the reduced equation of state and write 

L+j\q*-V) = %0, <u.i) 

where n, 9 and are the reduced pressure, volume and temperature respec- 
tively. We may expand the liquid from D to B (see figure 70) by applying a 
hydrostatic tension. At B the liquid would become a vapour, that is it would 
pull apart. The pressure at which this occurs may be obtained by writing 
equation (11.1) as 

80 3 (11.2) 

(3*-l) *»' 

and determining the minimum at B by putting — = 0. This gives 


194 Gases, liquids and solids 

Figure 70. The tensile strength of a liquid treated as an extreme case of a van der 
Waals fluid. Using the reduced equation of state it is shown that the liquid fails 
when a negative pressure n m is applied. 


3-80 / 6 \ „ 

This defines the value of I at B; we call this j m . Then 


If we insert this in equation (11.2) we obtain the corresponding value of n 
which we call it m . 

it m = 



We see from this equation that n m = when 4 m = § . This is the 'last' iso- 
therm for which the liquid has no tensile strength; it occurs when ^ = § or 
V = § V c . If the liquid is less condensed than this it can exist only if a positive 
pressure is applied. 

At lower temperatures — n becomes larger. It reaches its maximum (negative) 
value, as equation (11.3) shows, when = 0, that is when 3^ = 1. Putting 
^ = \ in equation (11.4) we obtain as the maximum value 

!- 2 

The limiting hydrostatic tension is then 

195 The liquid state 


For water P e = 218 atmospheres, so that 

P = - 5600 atmospheres. (11.6) 

Before we leave this calculation we note that = corresponds to absolute 
zero temperature and * = £ corresponds to V = \V C = b, that is the mole- 
cules are packed as close as the van der Waals model will allow. We may ask 
how such high negative pressures can be meaningful in a liquid. Why does the 
vapour not form when the smallest negative pressure is applied? The answer 
is that the nucleation of a vapour bubble is very difficult. For example to 
open up a spherical hole of radius r in a liquid of free surface energy y, the 
vapour pressure would have to exceed 

/>=-• (11.7) 

For water y = 72 erg cm. -2 and assuming r is the size of a water molecule 
( = about 3 A.) we obtain 

p = 4800 atmospheres. (11.8) 

Generally the vapour pressure is a negligible fraction of this value. However, 
we could produce the nucleation of bubbles, even if the vapour pressure is 
very small, by applying an external hydrostatic tension of 4800 atmospheres. 
Cavities would form and the liquid would then rupture. This tension is very 
close to that calculated in equation (11.6). 

Readers may note that we could, in fact, estimate the breaking strength of 
the liquid simply in terms of the surface energy. Consider a liquid column of 
unit cross section. If we pull it apart we create two unit areas of surface, so 
that the work done is 2y. The intermolecular forces are short range and if they 
become negligible for a distance greater than say x, the force F to pull the 
column apart will be of order Fx x = 2y or F = 2y/x. If x is about 5 A. we 
again obtain as the breaking stress (force per unit area) a value close to that 
given in equations (11.6) and (11.8). Exact agreement is not to be expected; 
the main point we wish to make is that because of intermolecular forces 
liquids can. in theory, withstand large negative pressures. In practice, of course, 
liquids are generally much weaker because of dissolved gases which readily 
nucleate cavities as the negative tension is applied. This is particularly marked 
at the walls of the vessel where small amounts of gas may be trapped in the 
surface irregularities. Carefully outgassed liquids in smooth-walled containers 
can, however, possess tensile strengths up to one tenth of the values calculated 

11-2 The structure of liquids: the 'radial distribution function' 

Before turning to a consideration of a liquid as a modified solid we mention 
one characteristic feature of a liquid which distinguishes it in a very basic way 

196 Gases, liquids and solids 

from a gas. A gas has no structure. If we study the diffraction of neutrons, 
electrons or X-rays by a monatomic gas we find that there is no coherent 
diffraction; the gas behaves like a diffraction grating in which the spacings 
between the lines are random. (For diatomic molecules one obtains diffraction 
maxima corresponding to the mean distance between the atoms in the mole- 
cule.) If, however, a monatomic gas is liquefied it shows marked coherent 
diffraction; the maxima are not sharp as with a solid crystal, but are diffuse. 
The information obtained by such diffraction experiments is presented most 
graphically by means of the 'radial distribution function' which may be derived 
in the following way. We choose a molecule at random within the liquid and 
draw a series of concentric spheres around it such that the volume interval 
between two neighbouring spheres is a constant. The distribution function is 
the average number of molecules contained between each spherical shell. It is 
evident that the radial distribution function is a measure of the average density 
as a function of distance from some arbitrary origin. A typical result from 
X-ray diffraction is given in figure 71 and similar results are obtained from 

-? 3 


liquid Hg (normalized height units) 
solid Hg (arbitrary height units) 

Figure 71. Radial distribution function plotted against radius measured from an 
arbitrary atom chosen as origin, (a) Liquid mercury (b) solid mercury - from 
results given by A. Guinier, X-ray Diffraction in Crystals, imperfect Crystals and 
Amorphous Bodies. Freeman. San Francisco. 1 963. p. 70. figure 39 For the liquid 
the vertical ordinate can be normalized such that for large distances the radial 
distribution function tends to unity (uniform density) For the solid this is not 
possible since in principle each peak is infinitely narrow and infinitely high. 
Instead the vertical ordinate has been drawn proportional to the number of atoms 
at the specified radial distance. It is seen that with the liquid the structure is lost 
beyond distances of 1 0A ; with the solid it persists indefinitely. 

the diffraction of neutrons. As a matter of interest the radial distribution 
function for the crystalline solid is also shown; this brings out the marked 

197 The liquid state 

difference between the liquid and the solid state. The solid behaves like a 
diffraction grating containing a very large number of regularly spaced lines: 
the liquid resembles the diffraction of light by a grating containing only a few 
lines. This suggests that there is short-range order in a liquid; any small group 
of atoms or molecules is in fairly regular array and resembles a crystal lattice 
but it is just a little out of register in a random way so that, as soon as one 
moves a short distance from the group, the neighbouring atoms (or molecules) 
no longer possess the long-range order characteristic of the solid crystal. 

The contrast between the liquid and solid states may be expressed in another 
way (see J. D. Bernal, Scientific American, no. 203, 1960, pp. 124-34, and 
Proceedings of the Royal Society, A280, 1962, pp. 299-322). Neutron diff- 
raction experiments suggest that a molecule in a liquid has time to vibrate 
10 to 100 times before the local structure changes. During that time the struc- 
ture of a liquid is physically, though not geometrically, similar to that of a 
crystal. The difference, however, is crucial. For example, the solid crystal, 
because of its high degree of order, admits only a limited degree of variation 
of composition. Crystallization as a means of purifying substances is evidence 
of the particular architecture that marks each crystalline phase. By contrast, 
liquids mix much more readily either with other liquids or with solids or with 
gases (though of course they do not mix as readily as gases which are always 
perfectly miscible with one another). We may describe this situation by saying 
that 'atoms in a liquid do not occupy such closely specified positions as they 
do in solids. They have more elbow room, and they are not so particular about 
their partners.' 

Nevertheless the structural resemblance to the solid over limited regions 
appears to be indisputable. This conclusion is supported from a completely 
different type of measurement. It is found that when metals melt their electrical 
resistance increases by a factor of only about two. As mentioned in a previous 
chapter the resistance is primarily due to non-coherent scattering of the 
electrons by the metal ions. This implies that, to the conductance electrons, 
the metal ions, in the liquid, appear to be oscillating about regular sites, very 
much as in the solid state. 

There are two main differences between the solid and liquid state. First, as 
discussed above, there is the absence of long-range order in liquids: in this 
sense they closely resemble amorphous solids, glasses and polymers. It is 
significant that these materials can flow viscously like liquids, although as 
pointed out in Chapter 7 their viscosity is generally enormous compared with 
that of true liquids. On the other hand, with crystalline solids flow or 'creep' 
can occur, but it takes place extremely slowly and through an entirely different 
mechanism - through the movement of dislocations, the behaviour of which 
is determined by the highly ordered structure of the material. This brings us 
to the second major difference: compared with the molecules in the solid state 
(both crystalline and amorphous), the molecules in liquids have very much 
greater mobility. They can swop places or shuffle around to other positions 
with relative ease. 

198 Gases, liquids and solids 

11*3 The liquid as a modffled solid 

When solids melt they expand generally by 5-15 per cent in volume. One way 
of describing this is to say that there is an increase in free volume between the 
constituent molecules or atoms. Another is as follows (cf. E. A. Moelwyn- 
Hughes, private communication; see also Physical Chemistry, Pergamon 
Press, 1960). If one thinks of a typical -crystalline solid for which each mole- 
cule, or atom, has say 8 or 12 nearest neighbours, we may regard the liquid 
state as a solid from which one nearest neighbour has been subtracted. This 
leaves the material with the right amount of volume expansion and a reason- 
able degree of short-range order. What is the nature of this disorder? The 
X-ray diffraction evidence suggests that the mean intermolecular distance 
does not change as much as would be expected from the bulk volume ex- 
pansion. For example, when solid argon melts the mean argon-argon spacing 
increases by only about I per cent, whereas the volume change would lead us 
to expect a mean increase in separation of about 5 per cent. Such evidence 
may be used to justify the view that the melting process consists mainly in 
producing 'holes' in a structure which otherwise closely resembles the solid 
structure. This idea was first developed by J. G. Kirkwood and constitutes the 
simplest form of the 'hole-theory' of liquids. As we shall see later the modi- 
fied 'hole-theory' introduced by H. Eyring provides a very convenient model 
for the study of viscous flow in liquids. 

Attempts have been made to treat the disordered structure and the increase 
in free volume in the liquid state in terms familiar to the solid-state physicist. 
One approach is to concentrate on the role of vacancies or holes (A. L. O. 
Rees, 1966, private communication). As the temperature rises the concen- 
tration of vacancies increases. So long as vacancies are individual holes the 
behaviour is still that of a solid. As soon, however, as two or three vacancies 
coalesce a whole aggregate of molecules surrounding the hole becomes 
relatively mobile. The material is now very much more like a liquid. One 
advantage of this approach is that it stresses the co-operative nature of the 
melting process. 

11-4 The liquid state 'std generis' 

The most difficult approach is to tackle the liquid state without directly 
referring to the gaseous or solid state. The simplest method is that due to 
J. D. Bernal and his colleagues in England and to G. D. Scott in Canada. It 
may be referred to as a random packing model. Recognizing the disordered 
but relatively close-packed nature of the liquid structure Bernal took a large 
number of plasticine spheres, chalked them to stop them sticking, placed 
them in a football bladder, removed the air to avoid bubbles and then squeezed 
them together until they filled the whole of the space. On examining the 
aggregate it is found that the spheres have become polyhedra of various 
irregular shapes. The most common number of faces is thirteen (this is not 
the true co-ordination number since the centres of some of the neighbours 

199 The liquid state 

are more than the average distance apart). More striking is the observation 
that the most common number of sides to a face is five. This is of great signi- 
ficance for the following reason. In crystallography molecules can be arranged 
with two-fold, three-fold, four-fold, or six-fold symmetry, never with five-fold 
symmetry. This is because "one cannot form regular patterns with five-fold 
symmetry that will fill space solidly and extend indefinitely in three dimensions. 
It is like trying to pave a floor with five-sided tiles'. This is illustrated in figure 
72 and shows how this type of arrangement would explain the presence of 
short-range order and the absence of long-range order. 

5 fold 6 fold 

Figure 72. The difference between five-fold symmetry and six-fold symmetry, 
after Bernal. 

In a later study Bernal emphasized the point that the attractive forces 
between molecules play little part in influencing the packing which occurs 
in the condensed state: the packing is almost entirely determined by the re- 
pulsive forces, that is the molecules behave very much like hard spheres. As 
Bernal remarks the key word (originally used in a different sense) is 'the one 
used by Humpty Dumpty in Alice Through the Looking Glass, impenetra- 
bility'. We are familiar with this in the solid state where, for practically all 
crystalline structures which are not dominated by covalent or other directed 
bondings, the arrangement is rather like that arising from packing the mole- 
cules as close as their repulsive forces will allow. For simple atoms, indeed, 
the crystal structure is like that of close-packed spheres. Bernal therefore 
considered the random packing, not of deformable plasticine spheres as in 
the earlier work, but the packing of a large number of hard steel balls. After 
introducing black paint and letting it harden the aggregate could be broken 
open and studied. In this way the number of contacts and near contacts 
could be determined. The model gives a co-ordination number varying 
between four and eleven, and Bernal considers this variation in contact 
number from molecule to molecule, to be the most significant feature of the 
irregular liquid structure. The packing very adequately reproduces the radial 

200 Gases, liquids and solids 

distribution function for liquid argon derived from neutron diffraction 
experiments (see figure 73), though this is probably not a very critical test. 



0.8 1.2 1.6 2 

distance from centre in units of sphere diameter 

— _ from neutron diffraction 

V • calculated from random packing model 

Figure 73. Radial distribution function for liquid argon plotted against distance 
from centre in units of sphere diameter Results from neutron diffraction. O V cal- 
culated from random packing model (• Bernal, V Scott). 

Further, the model does not allow vacancies between neighbours to be large 
enough to admit a third molecule. 

The Bernal model aims to provide essentially an instantaneous snapshot 
(in say 10~ ,s sec.) of the structure of a liquid. If we now add to this the reality 
of thermal motion we see that the molecules are continuously changing the 
number of nearest neighbours, continuously shifting and shuffling around, 
and occasionally are able to allow one molecule to squeeze its way between 
neighbours without leaving a vacant hole. This aspect of the model may be 
open to criticism. For example, with a solid, because it is rigid, vacant sites 
or holes can exist in the lattice. The concentration of holes increases as the 
temperature is raised and by the time the solid reaches its melting point (but 
is not yet melted) about one site in 10* or 10 s is vacant. Once melting occurs 
there is a further increase in volume, but according to Bernal the free space is 
more uniformly spread out. The X-ray evidence quoted above suggests that 
this is not possible, even at temperatures just above the melting point; in any 
case the thermal expansion of liquids is rather large so that at somewhat 
higher temperatures there must be vacancies or holes. Indeed, as Bernal him- 
self points out, it is this increase in holes and the consequent drop in average 
co-ordination number that marks the transition from the coherent liquid 
phase to the incoherent gas phase (see figure 74). Bernal suggests that this 
probably occurs when the co-ordination number has fallen to an average 
value of three or four. Since the co-ordination number in the solid state (or in 

201 The liquid state 



coherent liquid \ incoherent gas 

S!2*L_„ J< , ^ H, 

packed expanded associated free 

Figure 74. The transitions from the close-packed coherent liquid phase to the 
expanded, associated and finally free (incoherent gas) phase, as proposed by 

the liquid state near the freezing point) is about twelve, this means that the 
density of a gas at its critical point is between one quarter and one third of 
the density of the most highly condensed state. This is in fairly good agree- 
ment with observation for most normal liquids. 

The Bernal approach is essentially that of a crystallographer; his model is a 
static one and concentrates on structural features. A far more analytical and 
difficult approach is that due to M. Born and H. S. Green. They carried out 
an extensive study of the mutual potential between molecules in the liquid 
state. They treat the molecules first in pairs, then in triplets and so on. The 
treatment is extremely complicated and the results are not easy to apply. In a 
sense their approach approximates to the theory of dense gases. A calculation 
is made of sets of co-ordinates of random collections of hard spheres. As the 
density of spheres is increased the behaviour passes from that of a gas to that 
of a dense gas and then to some other state. In this condition, by an iterative 
process, the co-ordinate positions can be repeatedly perturbed until a state 
of minimum energy is obtained. This corresponds over a critical density range, 
to the liquid state. The computation here is formidable even for only 100 

Yet another approach is that associated with the name of B. J. Alder; it is 
sometimes referred to as the method of molecular dynamics. It involves ela- 
borate computations but the basic concept is simple. The molecules are 
treated as hard smooth perfectly elastic spheres, although as a later refinement 
an attractive square-well potential with a rigid core has been used. The mole- 
cules are placed in a box and started off with equal speeds in random directions 
and their subsequent motion determined by applying Newton's laws of 
motion. The boundary conditions chosen are such that a particle which passes 
out of one side of the box re-enters with the same velocity through the 
opposite side. In this way the number of molecules in the box and the total 
energy remain constant. The collisions within the system are analysed by a 
computer. It is found that after relatively few collisions the velocity distri- 

202 Gases, liquids and solids 

button becomes very nearly Maxwellian. The total energy is determined by 
the initial velocities and of course remains constant ; it determines the effective 
temperature of the system. 

The behaviour depends on the velocity and on the packing of the particles. 
At one extreme the particles oscillate about an equilibrium position; this 
corresponds to the solid state. At the other extreme the particles move around 
throughout the box with relative freedom; this corresponds to the vapour 
regime. At some intermediate stage the particles vibrate about equilibrium 
positions but are also able to change places and slowly diffuse from one region 
to another. This corresponds to the liquid state. 

We quote some typical results obtained by Alder and Wainwright (B. J. 
Alder and T. E. Wainwright, NuoVo Cimento, volume 9, 1958, supplement 1, 
p. 16 and, Journal of Chemical Physics, volume 31, 1959, p. 459), for 32 hard 
smooth particles in a box. They were able to map out a view of the calculated 
particle trajectories as seen from one face of the box. If v is the volume occu- 
pied by the particles and v the volume they would occupy in a close-packed 
situation, they observed a very interesting instability for a ratio v/v = 1 525. 
For one group of collisions the particle trajectories appear as shown in figure 
75(a) and calculation shows that the pressure on the walls is small; clearly 

(a) (b) 

iff ^ ■*&-»-* y^^m* 

k .***% 

t4P%-4t«l htg 

ft ,.kky <& & tf UL---^-?8S^x*^ 

Figure 75. End face view of box showing calculated particle trajectories for 32 
hard elastic spheres (zero attractive forces) for a packing VI V = 1-525. (a) Behavi- 
our resembling a solid, (b) Behaviour resembling a liquid. The particles spend most 
of their time oscillating within a cage of other particles, but are also able to swop 
places. See particles A and B ; C, D and E. 

this represents the solid state. In other groups of collisions the particle tra- 
jectories appear as in figure 75(b) and calculations show that the pressure on 
the walls is large. This is interpreted as the liquid phase. It is interesting to 
note that in the 'liquid* state the particles oscillate about a mean position but 
occasionally swop places (see figure 75(6), particles A and B, particles C, D 

203 The liquid state 

and E). The situation shown in these figures refers to only 32 particles in the 
box. Apparently the system is not sufficiently large for the solid and liquid 
phases to exist in equilibrium: the system is in just that density range where 
pan of the time it is in the liquid phase and part of the time in the solid phase. 

Although the results reproduced in figure 75 are very striking, some reserve 
is necessary since an assembly of particles which show no attraction for one 
another can hardly be considered as a very realistic model of a liquid or a 
solid. However, from the point of view of packing, the structure scarcely 
depends on intermolecular forces (see Bernal above). 

If the molecules are less closely packed the results obtained are as shown 
in figure 76. Figure 76(a) shows the potential for particles with a square-well 




Figure 76. (a) Potential diagram of spheres possessing a square-well attractive 
field and a hard elastic core. The repulsive core has an effective diameter o t . 
(b) End face of box showing calculated trajectories of 108 such spheres. The 
behaviour of the particles resembles the vapour phase. 

attractive field whilst figure 76(6) shows the particle trajectories for 108 such 
particles in a box ; the behaviour evidently represents the liquid-vapour region. 
The Alder model brings out very clearly the dynamic features of the condensed 
state and the similarities and differences between the solid and liquid phases. 

1 1 -5 Ice and water 

We introduce here a short digression on the relation between ice and water. 
Water, which is the basis of all our ideas of liquidity, is in a sense an atypical 
liquid. The forces between the molecules are very strongly directional; they 
arise from the interaction between electron-deficient hydrogen atoms in one 

204 Gases, liquids and solids 

molecule and electron-rich oxygen atoms in another. In the solid state this 
hydrogen bonding leads to a very open structure in which every water mole- 
cule is surrounded symmetrically (in the form of a regular tetrahedron) by 
four other water molecules. Consequently in ice the co-ordination number is 
four. As the temperature is raised the co-ordination number remains constant 
but the mean distance between the molecules increases. There is thus a small 

• water • 

density p=k~i 

slopes exaggerated 



1 2.8- 


molecular separation r 





coordination number n 



temperature °C. 

Figure 77. The melting of ice. Schematic diagram showing (a) the change in co- 
ordination number. (/») the change in molecular separation, (c) the corresponding 
change in density as ice is heated from — 5"C. to+15"C. 

decrease in density with increasing temperature as with all 'normal' solids. 
At the melting point, however, a radical change occurs. The thermal energy 
is sufficient to destroy the regular bonding and the co-ordination number 
actually increases. At 0°C. the average co-ordination number appears to be 
between 4-5 and 5; with rising temperature this probably increases but soon 
reaches a constant value. This increase on its own must lead to a correspond- 
ing rise in density. However, an increase in temperature also leads to an 
overall increase in the mean distance between the molecules (thermal expan- 
sion). This expansion, on its own, would produce a decrease in density with 
increasing temperature. When combined with the change in co-ordination 
number it results in a 10 per cent jump in density on melting, a maximum 

205 The liquid state 

density at 4°C, followed by a steady decrease in density at temperatures above 
about 10°C. The behaviour is shown schematically in figure 77. The details are 
by no means definite but the general pattern is reasonably valid. Another way 
of looking at this is in terms of the 'mixture-models' of water (e.g. H. S. Frank 
models). These stress the idea that water may not simply be a random or an 
ordered arrangement of individual molecules with uniform characteristics 
and bonding. On the contrary water may consist of two or more different 
molecular species, each species being structurally organized in its own dis- 
tinctive way, with its own degree of co-ordination. The effect of temperature 
on this model is to increase the proportion of one species at the expense of 
the other. Whichever approach is adopted, the main point brought out by this 
discussion is the immense importance of the co-ordination number in deter- 
mining the detailed properties of liquids. 

11-6 Genera] approach 

In this chapter we shall treat the liquid as a modified solid, particularly when 
we are concerned with energies. When we consider flow we shall treat the 
liquid essentially as a solid, the molecules of which have a fair degree of 
mobility because of the presence of holes or free volume. This is the Eyring 
theory of viscous flow in a liquid; it differs from the Bernal theory in concen- 
trating on thermal fluctuations rather than on the structural nature of the 
liquid state. It is much closer to the Alder model of the liquid state. 

Before leaving this discussion we may make the following point. In the 
gaseous state the molecules are, more or less, completely independent. In the 
solid state they interact but the interaction, at least as far as crystalline solids 
are concerned, is capable of analysis precisely because of the long-range order 
which obtains. Solids are indeed an example of highly ordered co-operative 
interaction. By contrast the behaviour of a liquid is the result of co-operative 
interaction within a system of random order, or, at best, low-range order. 
It is this feature which makes the theory of liquids so difficult. 

1 1 -7 Latent heat of fusion 

The latent heat of fusion is between one tenth and one thirtieth of the latent 
heat of sublimation. The latter represents the complete disintegration of the 
solid into the gas. Thus the change from solid to liquid involves a relatively 
small energy increase compared with the change from liquid to gas. For all 
the elements the latent heat of fusion is very small; for most compounds the 
latent heat of fusion corresponds approximately to the change from free 
vibration in the solid state to free rotation in the liquid. There is no detailed 
theory that is satisfactory. For many liquids the specific heat is of the order 
3-6 cal. mole" 1 "C." 1 

206 Gases, liquids and solids 

11-8 Melting point 

The most astonishing feature of fusion is that the melting point is so sharp. 
If melting were always accompanied by a volume increase we could under- 
stand (if not explain) the sharpness of the melting point in the following way. 
Melting represents essentially the 'loosening-up* of the solid structure. If 
local melting occurs molecules on the edge of melted regions are exposed to 
molecules which are already a little further apart than usual. It would not be 
unreasonable to see in this the basis of a runaway situation where a little bit 
of melting soon engulfs the whole solid. If a pressure were applied to the 
solid to prevent a volume increase we should, in fact, find that the melting 
point is increased (Clausius-Clapeyron equation). Thus melting could cover 
a whole range of temperatures depending on the extent to which the volume 
increase is inhibited; only if the solid were completely free to expand would 
melting occur at a single sharply defined temperature. This 'explanation' 
would, of course, be a little embarrassing if we attempted to apply it to the 
melting of ice since here melting is accompanied by a decrease in volume. 

As we saw earlier it is not possible to explain melting in terms of our 
potential energy curve, since this can only describe basically the solid state 
or the gaseous state. However, Lindemann (1910) suggested that one could 
obtain a good idea of the melting point from the P.E. curve by assuming 
that melting occurs when the amplitude of vibration exceeds a critical fraction 
of the atomic spacing. At this stage numerous collisions occur between 
neighbours and the crystal structure is destroyed. 

Let x be the vibration-amplitude when melting occurs. If /is the force 
constant between one atom (or molecule) and its neighbour and the motion is 
simple harmonic the mean total energy during vibration will be 

u. = iA§. (11.9) 

Since melting points are generally high we may assume the classical value for 
the internal energy, namely kT per atom or molecule for each principal 
direction of vibration. Thus at the melting point we write 

«r = kT m = \fx%. (11.10) 

But as we saw in an earlier chapter, for a simple cubic structure,/ = Ea where 
a is the atomic (or molecular) spacing and E is Young's modulus of the solid. 

Hence kT m = \Eax%. 

We assume that melting occurs when x is some fraction of the atomic 
spacing a. Let x be fia where < fi < 1. 

T Ea3fi2 = £g3jV ^ 1 = EV P* 
2k 2R 2R 

where V = molecular volume, assuming a cubic structure. 

r. = ^x^^, ( 

2p R 
207 The liquid state 

where p = density of the solid. 

Some typical results are given in Table 25 where it is assumed that 2 has 
the value of ^ j e ^ Xo ~ ? . 

Table 25 Melting point T m (°K.) of some solids 


Edyne cm.' 1 

p g. cm.' 2 


T m 'K. 









8-3 x 10" 

























70 xlO" 





This model has assumed simple cubic packing (so that Na 3 = V) and fi a 
constant fractional amplitude of vibration at melting. Considering the crudity 
of the model and the range of materials to which it has been applied the 
agreement is surprisingly good. 

11-9 Vapour pressure 

11-9-1 Molecular interpretation of vapour pressure 

Suppose we place a quantity of liquid in a sealed container and evacuate the 
air above it. Some of the liquid will evaporate since a certain fraction of the 
molecules will have enough excess thermal energy to escape from the attrac- 
tion of neighbours. Some of these will return and condense. Equilibrium will 
be reached when the rate of evaporation equals the rate of condensation. The 
pressure exerted by the vapour under these conditions is known as the satura- 
tion vapour pressure or more usually the vapour pressure. It increases as the 
temperature is raised. The equilibrium vapour pressure will be almost identical 
in the presence of air or another gas (see however the section below on the 
effect of external pressure), since the vapour and the alien gas exert then- 
partial pressures independently. 

If the pressure above the liquid is set at some specified level, e.g. 1 atmo- 
sphere, the vapour pressure may be increased by raising the temperature of the 
liquid, until it is equal to the external pressure. At this point the liquid can 
evaporate freely: bubbles of vapour form copiously within the bulk of the 

208 Gases, liquids and solid 

liquid and the liquid is said to be boiling. The boiling point is thus the tem- 
perature at which the vapour pressure equals the external pressure - usually 
one atmosphere. If the external pressure is reduced the boiling point is lowered. 
It is a matter of experience that the boiling point on the absolute scale, at 
atmospheric pressure, is approximately two-thirds of the critical temperature, 
but there is no simple explanation for this. 

The molecular interpretation of vapour pressure is simple. The molecules 
in the liquid have a wide range of thermal energies. The only molecules which 
can escape are those whose thermal energy is sufficient to overcome the 
attraction of neighbouring molecules, if the latent heat of vaporization per 
gram-molecule is L, the energy for a molecule to escape is 

e = -£ . (11.12) 


It follows that the average potential energy of the molecules in the vapour 
phase is greater than the mean value in the liquid phase by an amount e. 
Consequently if n v and n L represent the number of molecules per unit volume 
in the vapour and liquid phases respectively and if the Boltzmann distribution 
holds for the liquid as well as for the vapour, we have 


- exp ("£) - exp (-]£r) " exp ("£)• (U13) 

As the temperature is raised the ratio n Y \n L increases but n L scarcely changes 
with temperature. Since the vapour pressure p is directly proportional to the 
number of molecules per unit volume n v in the vapour phase we may rewrite 
equation (11.13) as 

p = Aexp(~\, (11.13a) 

where A is a suitable constant. This relation is almost exactly correct. 

We may approach the vapour pressure relation in another way. Suppose 
we consider the liquid to resemble a very dense gas containing n L molecules 
per c.c. The number per c.c. with velocities, normal to the surface, between 
u and u+du is given by equation (5.41) in Chapter 5. The number striking 
each sq. cm. of surface per sec. is then 

Udn = nL {^TTJ j} XP (-WT) UdU - 

If we now specify that of these molecules only those with energy greater than 
s (where e = LIN ) can escape, the number per sec. crossing each sq. cm. 
of surface into the vapour phase is the same integral as above but the limits 
are from mu 2 l2 = s to oo. This gives us the number escaping per sq. cm. 
per sec., 

209 The liquid state 

But if n v represents the number of molecules per c.c. in the vapour phase, the 
number striking each sq. cm. of surface per second will be (see Chapter 5, 

ennatinn ^iH) 

equation (5.41)) 

n v (—Y 

If all those molecules striking the liquid surface are adsorbed (this is the 
most dubious part of the treatment) then equilibrium occurs when the number 
escaping from the liquid exactly equals the number arriving from the vapour 
phase, that is 


niexp r^r MF - 

This is identical with equation (11.13). 

1 1 -9 -2 Effect of temperature on vapour pressure 

We use equation (11.13) to calculate the effect of temperature on vapour 
pressure. Since 

we have by differentiating 

dp . L [ L\ L L p 

Tt= a W^ [-rt) - rT>* p = Rj x f (11 - 14) 

If the vapour behaves like a perfect gas and V is the volume occupied by a 
gram-molecule of vapour at Twe havepK = RT, so that 


p = -y- ai.i5> 

Equation (11.14) becomes 
dp L 1 

The correct relation, according to thermodynamics, is 

dp _ L 1 

dT~ T(V-V e )' (1W7) 

where V. is the volume occupied by a gram-molecule of the liquid. This is 
210 Gases, liquids and solids 

known as the Clausius-Clapeyron equation. We may note that a plot of log p 
against T~ ' should give a straight line of slope equal to L/R. This is indeed 
found to be so. 

Effect of external pressure on vapour pressure 

If we apply an external pressure, for example by compressing with an inert 
gas, how is the vapour pressure p of the liquid affected? The answer is sur- 
prising; p is increased. This may be explained in terms of figure 78 which 









Figure 78. Effect of external pressure on the vapour pressure of a liquid, (a) Initial 
condition - the membrane is permeable only to the vapour, (b) Condition when 
external pressure x is applied, (c) Alternative model. 

represents an idealized experiment. The top horizontal limb of the tube abed 
contains a membrane permeable only to the vapour above the liquid AB. 
When the space above the liquid contains only vapour, A is level with B. If 
now an inert gas is pumped into the right-hand limb at a pressure n, since it 
cannot pass through the membrane, the level A rises to A' and B falls to B'. 
If A is the difference in height, clearly 



Since the membrane is permeable to the vapour, the vapour over A' must be 
in equilibrium with the vapour over B'. This implies that the vapour pressure 
at B' is greater than over A' by hpvg- The increase in vapour pressure is thus 

A/> = hp v g = n — . 


Another way of understanding this is to consider the arrangement in figure 
78(c). The liquid is contained in a cylinder one end of which is closed by a 
movable piston. The bulk movement of the liquid is blocked by a membrane, 
which is impermeable to the liquid, but permeable to the vapour. The whole 

211 The liquid state 

is maintained at a temperature T, so that heat is available for evaporation. 
If the piston is compressed with a pressure n until one gram-molecule of 1 i quid 
has been forced to evaporate the liquid diminishes in volume by amount V L ; 
the piston does work nV L on the liquid. Consequently the thermal energy 
necessary for evaporation is reduced from L to L - n V L . The vapour pressure 
then becomes 

'-M-^HM^ME) <■"• 
-*(■♦£'♦ -)-* +l £'- 


Ap =p-p = — — . (UM) 

If the vapour behaves like a perfect gas p V r = RT so that we may insert 

Po = — in equation (11.21). This gives 


A V L py 

Ap = »— = » — . (11.22) 

y r Pl 

Since at room temperature pvlp L is of the order of 10" 3 , the increase in vapour 
pressure with external pressure is generally small. Descriptively one might say 
that the external pressure squeezes out the liquid molecules. It is not so easy 
to see why a negative pressure reduces the ease of escape of the liquid molecule. 
Nevertheless it is clear from the preceding experiment that a negative pressure 
will in practice reduce the vapour pressure. 

11-10 Surface tension 

11-10*1 Surface energy 

It is common experience that liquids tend to draw up into drops; small drops 
form perfect spheres. Since a sphere is the geometric form which has the 
smallest surface area for a given volume we conclude that the surface of the 
liquid has a higher energy than the bulk; the liquid is therefore always 
attempting to attain its lowest energy state by reducing its surface area. This 
energy excess is called the free surface energy y and is measured in erg cm.~- 
If a liquid has area A its free surface energy is yA. If we increase the surface 
by doing work on it the work done is 

d(yA) A by 

Ha= 7+a Ta' <«■*> 

212 Gases, liquids and solids 

With a liquid, however much we stretch the surface the surface structure 

remains unchanged so that — = 0. This is in contrast with a solid. 


Work done per unit area of extension = y. 


Recent experiments have shown that very thin films of water have the same 
surface energy as two surfaces separated by bulk water. The identity holds 
down to a thickness of only 20 A. This implies that the molecular forces res- 
ponsible for the surface energy (see below) are very short range. 

The free surface energy is equivalent to a line tension acting in all directions 
parallel to the surface. Consider a liquid surface (as shown in figure 79) of 

Figure 79. Sketch showing equivalence of surface energy per unit area and line 
tension per unit length. 

width L and length X, Suppose at the barrier AB we apply a force F parallel 
to the surface and normal to AB so that we extend the length of the surface 
by an amount x. The area increase is Lx and the work done in increasing the 
surface area is Lxy. This must be equal to the external work done Fx. 

Hence Fx = Lxy, 


l =y ' 


Thus the surface energy is equivalent to a surface tension or line tension 
y dyne cm. -1 

Molecular interpretation 

The free surface energy of a liquid lends itself to a very simple molecular 
interpretation. Molecules in the bulk are subjected to attraction by surround- 
ing molecules; the field is symmetrical and has no net effect. At the surface, 
however, the surface molecules are pulled in towards the bulk of the liquid, 
Apart from a few vapour molecules there is no attraction in the opposite 
direction. Consequently if we wish to increase the surface area we have to 
pull molecules up to the surface from the bulk against this one-sided attrac- 
tion. This accounts for the surface energy. 

213 The liquid state 

Alternatively we may say that in moving molecules from the bulk towards 
the surface we are continuously breaking and reforming bonds. At the top, 
however, roughly half the bonds are not reformed. Since bond formation 
implies a decrease in energy the incomplete reformation of bonds at the sur- 
face means that the surface has a higher energy than the bulk. Practically the 
whole of this process is confined to the last few molecular layers and it is for 
this reason that extremely thin liquid films have the same surface energy as 
'bulk' liquid (see above). Without going into further details it is evident that 
the surface energy is of the order of one half the energy required to break all 
bonds, i.e., one half the latent heat of vaporization of a molecular layer. This 
is essentially the same as the model we discussed earlier for solids and we 
emphasize again that the intermolecular forces responsible for the surface 
tension of a liquid are of the same nature as those responsible for the pressure 
defect of the corresponding gas. 

If L is the latent heat of vaporization per mole and A/the molecular weight 
and p the density, we have, as an approximate relation [cf. equation (7.9)], 

2N \ 


For liquid argon this gives a value of 22 erg cm. -2 for the surface tension 
compared with the observed value of 13; for liquid neon 6 compared with 
5-5; for nitrogen 18 compared with 10-5; for oxygen 22 compared with 18; 
for benzene 180 compared with 40 and for mercury 1000 compared with 
600 erg cm. ~ 2 

Thermodynamics provides another quantity, the total surface energy h. 
When a surface is increased in area, apart from the surface energy term y, 
there is also a 'latent heat' term. Heat must be provided to keep the tempera- 
ture constant. Its magnitude is — T — . Hence 'the total surface energy' is 


h=y-T^,. (11.27) 

Equation (11.26) does not distinguish between these two concepts; it is really 
a model for y at 0°K. so that the poor agreement with the room temperature 
experimental values must not be taken too seriously. On the other hand the 
difference between ft and y is often significant. For example, for water y = 72 
and h = 1 1 8 erg cm." 2 If fine droplets of water are allowed to coalesce so as 
to destroy their surface area the temperature rise is determined by h, not by y. 
This provides a very ingenious method of determining areas of fine particles. 
They are first equilibriated with water vapour so that their surface is fully 
covered with a condensed film of water only a few angstroms thick. They are 
then immersed in water in a delicate calorimeter. If the heat given out is AQ 
erg, the surface area is given by AQ/118 cm. 2 

214 Gases, liquids and solids 

Contact angle and contact equilibrium 

It is generally found that liquids with low surface tension readily wet most 
solids, giving a contact angle of zero; those with high surface tension often 
show a finite contact angle. In molecular terms we may say that if the cohesion 
between the molecules of the liquid is greater than the adhesion between the 
liquid and solid, the liquid will not 'want' to wet the solid, consequently it 
will show a finite contact angle. 
Contact equilibrium is shown in figure 80. The free surface of the liquid has 

Figure 80. (a) Contact angle equilibrium. (b) Contact on a thick specimen, 
(c) Contact on a thin mica sheet where the vertical component of the tension 
distorts the shape of the sheet. 

a surface energy y L ; the solid liquid interface a surface energy y SL and the 
exposed portion of the solid adjacent to the liquid where vapour has been 
adsorbed a surface energy y SF . If we consider a virtual movement of the con- 
tact region so that an additional sq. cm. of solid is wetted we have 

surface energy increase at solid-liquid interface = y S t, 

surface energy decrease at solid-vapour interface = y sr , 

surface energy increase of water surface = y L cos 9. 

By the principle of virtual work 

Vsv= ?st+7LCose. (11.28) 

This equation, which was first derived by T. Young, without proof, in 
1805 and by A. Dupre in 1869, is exactly what one would obtain if we took 
the horizontal components of the surface tension forces. This at once raises 
the question: what has happened to the vertical component y L sin 01 The 
answer is simple. The force is very small and its effect on a solid is negligible. 
If, however, the solid consists of a very thin flexible sheet, e.g., of mica, the 
vertical forces may be sufficient to distort the surface visibly. One is then aware 
of the reality of the vertical component. 

Equation (11.28) has been studied experimentally by Miss S. Kay, using the 
bifurcation of mica as a means of determining directly the surface energies 
of the solid. In the first experiment she determined y for a mica surface exposed 
to water vapour; this gives ysr. In the second experiment she determined y 

215 The liquid state 

when the surfaces were immersed in water; this gives y S t- Since water com- 
pletely wets mica (0 = 0) we should expect to find, from equation (11.28) 

7l = Vsv- ?sl. (11.29) 

Results obtained with water are given in Table 26. The table includes results 
for hexane which also wets mica. 

Table 26 Surface energy of mica in presence of vapour and liquid 




erg cm.~ 2 - 











The agreement is very satisfactory. 

11-10-4 Pressure difference across a liquid-vapour interface 

We shall now discuss a few of the general properties of capillarity which are 
shown by liquids. Before doing so we derive a simple relation for the pressure 
difference across a curved liquid interface. 

Let ABCD be a portion of a surface marking the boundary between air 
below and liquid above (see figure 81). Over a small region a curved surface 


81 . Pressure difference across 
a liquid-air interface. 

216 Solids, liquids and gases 

can always be described in terms of two radii of curvature in mutually ortho- 
gonal planes. These are known as the principal radii of curvature R, and R 2 . 
Suppose we assume that the pressure on the air side is p t and on the liquid side 
is p 2 . We now apply the principle of virtual work displacing the surface 
parallel to itself through a distance dx. Then Afi increases in length by 

dxBi = dx — ; similarly AD increases in length by dx 6 2 = dx — .Thein- 
#1 R 2 

crease in surface area is then 

Aa= (aB+AB — \ /bC+BC — \-ABxBQ 

A« = ABxBCxdxx (—+—). (11.30) 

This involves an increase in surface energy of yAa. The work done by the 
pressure difference px —p 2 is 

iPi -Pz) AB x AD x dx = yAa. 

The pressure on the concave side is greater than on the convex side of a liquid 
interface, by an amount depending on the surface tension and on the curva- 
ture. This relation becomes clearer in the following two very simple cases. 
For a spherical bubble of radius R within a liquid the surface area is AnR 1 . 
If a small expansion occurs increasing R by dR the increase in area of surface 
is %nRdR. The work that must be done by the pressure in the bubble is 
4nR 2 pdR. Equating this to the increase in surface energy we have 

4nR 2 p dR = SnR dRy, 

P--£. (" -32) 

This is exactly the result of equation (11.31) if R t = R 2 . The second case is 
for a soap bubble. Here two surfaces are involved and the pressure inside 
the bubble is 

*> = ^'. (1133) 

where y, is the surface tension of the soap solution. 

Capillary rise 

Suppose a uniform tube of small radius R is held vertically and lowered into 
a liquid of density p L which wets it (0 — 0). The liquid rises to a height h above 

217 The liquid state 

the common level. This is sometimes explained in terms of the surface forces 
around the periphery of the meniscus. The meniscus has peripheral length 
2nR and for zero contact angle the upward force is 2nRy. This is balanced 
by the downward weight of the liquid column hnR 2 p L g [figure 82(a)]. Thus 

force *-2nR7 


pressure defect "2-L 

- weight hTrR i p L g 





11 =hp L g 

Figure 82. (a) Simple calculation of capillary rise in terms of surface tension 
forces, (b) Inapplicability of method a to a conical capillary, (c) Calculation of 
capillary rise in terms of pressure defect 

f>PLg = 



Although equation (11.34) is correct, this is largely fortuitous. For example if 
the capillary was conical in shape [figure 82(fc)] the capillary rise would still 
be given by equation (11.34) where R is the capillary radius at the point where 
the meniscus formed. Clearly 2nRy could be considerably less than the weight 
of the column of liquid. 

The proper method is to treat the problem in terms of the pressure differ- 
ence across the meniscus. The meniscus for 8 = is a hemisphere of radius 
Ri = Ri = R- According to equation (11.31) the pressure at B will be less 
than at A by the amount 


But the pressure at A is the same as that at A' (it is atmospheric) and is the 
same as that at C except for a minute air pressure drop A'C. Thus the pressure 
in the liquid at B is less than the pressure in the liquid at C. Consequently the 

218 Gases, liquids and solids 

liquid is drawn op the tabe to a height h where 



We should obtain the same result for a conical capillary in agreement with 

experimental observation. 

This treatment also explains why the meniscus in a narrow capillary can 

be considered to be spherical in shape. If the capillary rise is large compared 

with the diameter of the tube, the height of rise is practically constant over 

the whole width of the capillary, i.e., the pressure defect hpig is constant. 

This means [see equation (11.35)] that the curvature of the meniscus must be 

constant. For a circular capillary the two principal radii of curvature must, 

by symmetry, be equal. Consequently the meniscus will be a portion of a sphere 

of almost constant radius, the value of which depends on the contact angle. 

For zero contact angle the shape will be hemispherical, the radius of curvature 

being equal to that of the tube R. If the contact angle is 6, it will be equal to 

R/cos 6 [see figure 83(a)] and the pressure defect giving rise to the capillary 

2 y cos G 
ascent will be equal to . As the capillary gets wider the height of the 



Figure 83. (a) Meniscus shape for large capillary rise must be of almost constant 
curvature, (b) For small capillary rise the radius of curvature becomes infinite 
at the centre. 

liquid column becomes small compared with the capillary diameter [see figure 
83(6)]. There is a very large difference between the pressure defect at L and at 
M. Indeed at M where the height of rise is negligible the curvature is almost 
zero {R — - oo) whilst it is finite at L and N. The detailed shape of the meniscus 

can be solved theoretically by equating the pressure defect yl — I — I at any 

\Ri Rif 
given point to the value of hpji at that point 

219 The liquid state 

11-10-6 Vapour pressure over a liquid meniscus 

We see from figure 82(c) that the vapour pressure over the liquid at A is less 
than over the liquid at C by hp y g where p v is the mean density of the vapour. 
We may write: 

vapour pressure defect 

= hp y g = hp L y. — y.g = ~x.-. (11.36) 

Pl R Pl 

This result is part of a more general statement: the vapour pressure over a 
concave meniscus surface is less than that over a flat liquid surface by the 
quantity given in equation (11.36). Similarly the vapour pressure over a con- 
vex meniscus is greater than over a flat surface by a similar amount. We may 
see this in a different way by considering the behaviour of a spherical liquid 
drop of radius R. The liquid within the drop is subjected to a surface tension 
pressure of amount p = 2y/Jt. As we saw in a previous section this increases 
the vapour pressure by the amount 

p ei = *i x ei. (1137) 

Pl & Pl 

11-11 Nucleation in condensation: the Wilson cloud chamber 

Consider the equilibrium between a vapour and its liquid and the problem of 
condensation. If there are already present a few droplets of radius of curvature 

r, the droplets exert a vapour pressure of — x — . This can be very large. For 

r Pl 
example with water vapour at 20°C, PvIpl is about 2 x 10" 5 . But if r is only 
5 A. the factor 2y/r has a value of about 3 x 10 9 so that the droplets are 
attempting to evaporate with a vapour pressure of about 6x 10* dyne cm. -2 
Since the saturated vapour pressure of water at 20"C. is about 2x10* dyne 
cm. -2 , the droplets will evaporate; they cannot grow by further condensation. 
There are only two ways by which condensation can be made to occur. One is 
to provide artificial nuclei with relatively large radii of curvature: this is, in 
practice, generally achieved by dust particles. The second is to produce super- 
saturation of the vapour, usually by sudden cooling. This is the basis of the 
Wilson cloud chamber. 

Consider a vessel closed with a moveable piston in which the vapour is in 
equilibrium with the liquid. Charged particles are projected through the 
vapour. Because the electrostatic energy of a point charge is lowered by the 
condensation of a dielectric on it, some condensation may occur. If the piston 
is now suddenly raised there is a marked adiabatic temperature drop. Although 
there is some volume increase the main effect is that the existing vapour pres- 
sure is much higher than the saturated vapour pressure at the new reduced 
temperature. Condensation occurs readily and the path of the particles is 
revealed as a series of liquid droplets. 

220 Gases, liquids and solids 

1112 Sup er h e a ti ng 

A similar problem is involved in the heating of a liquid to produce boiling. 
If the initial bubbles have a radius of curvature r, the vapour pressure in the 
bubbles must exceed 2y/r if the bubbles are to grow. For a bubble of radius 
r = 5 A, in a liquid such as water, this pressure is of the order of 3000 atmo- 
spheres. Thus unless the vapour pressure reaches this value the bubble will 
collapse. To obtain vapour pressures of this order the liquid must be very 
strongly superheated. With water, temperatures of the order of several 100°C. 
are required. Experiments with water from which dissolved gases have been 
eliminated do, in fact, show that bubbles will not form until temperatures of 
this order are reached. Of course, once the bubble begins to form at this high 
vapour pressure, the excess pressure necessary to produce bubble growth 
begins to decrease as r increases, so that the bubble grows at a catastrophic 
rate - the liquid explodes. 

Steady boiling at the true boiling temperature is achieved by providing 
suitable nuclei. These are often air pockets trapped at solid surfaces or gases 
dissolved in the liquid itself. 

11-13 The energy for capillary rise 

We raise here two points concerning capillary rise that are not often discussed 
in elementary textbooks. The first is, what is the source of the energy for 
capillary rise? Consider a tube of radius R. The liquid rises to a height h 
given by 

* -It*. < 1138 > 


The potential energy «i gained by the liquid is equivalent to the raising of a 

mass of liquid nR 2 hp through a height -. Hence 

e x = i nR 2 h 2 pg. (11.39) 

This energy comes from the wetting of the walls of the tube by the liquid' 
Descriptively we may see this in the following way. The surface energy of a 
solid arises from the asymmetric forces at the free surface. If we cover the 
surface with another material, e.g. a liquid, we reduce this asymmetry and 
hence reduce the surface energy. The energy liberated in this process is avail- 
able for pulling the liquid up the tube. We may make this quantitative as 

Consider a liquid with contact angle 0=0. The equilibrium condition, as 
we saw earlier, is 

?sv = 7l+ 7si. (H.40) 

If the liquid advances along the tube so as to cover 1 sq. cm. of the surface 

21 The liquid state 

we destroy one sq. cm. of ysv and gain one sq. cm. of y st .. There is no change 
in the area of the liquid meniscus. The energy given up by the system is thus 

ysv-Vsi.- (11.41) 

From equation (11.40) we see that this is equal to y L . Thus each sq. cm. of 
capillary surface wetted releases energy y L . The energy released for a rise of 
h is then 

s 2 = y L 2nRh. (11.42) 

Substituting from equation (11.38) for y L we have 

loss of surface energy e 2 = nR 2 h 2 pg. (11.43) 

This is exactly twice the gain in potential energy given by equation (11.39). 
This implies that if the liquid were non-viscous it would rise to a height 2h 
and oscillate between and 2k with its mean position at h. In practice vis- 
cosity consumes the excess energy very quickly. 

The behaviour is like that of a vertical spiral spring on to the end of which 
a weight is suddenly attached. The weight falls instantaneously to double its 
equilibrium depth and oscillates between this and zero until the kinetic energy 
is consumed by friction and static equilibrium is achieved. 

The second query is the following. Some solids are known to have much 
higher surface energies than others. For example if they are first thoroughly 
cleaned and then immersed in a liquid the amount of heat released may vary 
very widely; nevertheless if a liquid wets these surfaces (0 = 0), the capillary 
forces do not depend on the material of which the capillary is made; they 
depend only on the surface energy of the liquid. Why is this? The answer is 
implicit in equation (11.40). The adsorbed vapour on the solid will always 
adjust its thickness so that the quantity, 

ysv- ysL, 

is equal to y L (for = 0). The energy gain is thus dependent only on y L . We 
may describe this differently. The vapour adsorbed on the solid presents a 
surface to the advancing liquid which closely resembles the liquid itself. On a 
high-energy surface the adsorbed film will tend to be thicker than on a low- 
energy surface. In both cases (for = 0) the liquid is virtually wetting itself 
so that only y L is involved in the process. In this connection we may note 
that the range of forces is small and that the adsorbed film needs to be only 
10 to 20 A. thick in order to mask almost completely the influence of the 
underlying substrate. 

222 Gases, liquids and solids 

Chapter 12 

Liquids: their flow properties 

In this chapter we shall discuss the flow properties of liquids, and derive or 
state some of the more standard equations of flow. However, our main atten- 
tion will be directed to a molecular model of viscous flow; this follows closely 
the theory proposed by Eyring. 

12-1 Flow in ideal liquids: Bernoulli's equation 

As with gases we can assume the existence of ideal liquids in which internal 
forces play a trivial part, so that they have negligible surface tension or vis- 
cosity. Their flow properties are determined solely by their density. Of course, 
no liquids can have zero internal forces, they would not be liquid if this were 
so, but they can often behave as ideal liquids in flow if the inertial forces 

Consider the flow of such an idealized liquid and let us follow the path of 
any particle in it. If it moves in a continuous steady state we may draw a line 
such that the tangent at any point gives the direction of flow of the particle. 
Such lines are called streamlines. They are smooth continuous lines through- 
out the liquid and can never intersect - no fluid particles can flow across from 
one streamline to another. Let AB represent an imaginary tube in the liquid 
bounded by streamlines [figure 84(a)]. At A the liquid is at a height hi above 





Figure 84. (a) Streamlines of flow of a liquid in a gravitational field, (b) Flow of 
liquid from an orifice at a depth h below the free level. 

223 Liquids: their flow properties 

a reference level, the pressure acting at that point is p t , the cross-sectional 
area of the tube is at,, and the liquid flow velocity is i>i ; at B the corresponding 
quantities are h 2 , p 2 . <*%. o 2 . Consider the energy balance during a short time 
interval dr. The pressure p t at A drives a volume of liquid aiVidt through the 
tube: the work done is 

w t = pxtiVtdt. (12.1) 

The mass of liquid involved is pa^dt so that the kinetic energy it brings 
with it is 

v>i = iOxxiVtdtfr*, (12.2) 

and its potential energy is 

w 3 = (jxixVidfighi. (12.3) 

Since the motion is assumed steady and the liquid is ideal the same amount of 
energy leaves the tube at B. Hence 

(t?i + w t + w 3 )a. = (*>! + v>% + w 3 )». (12.4) 

Because the same mass of liquid leaves at B as enters at A, 

«iVtdt = at 2 v 2 dt. (12 .5) 

Equation (12.4) becomes 

- + iv 2 l + h 1 g = ?2 + lv 2 2 +h 2 g. (12.6) 

P P 

This is Bernoulli's equation of flow for an ideal liquid: the three terms on 
each side of the equation correspond respectively to the pressure energy to 
move the liquid, the kinetic energy and the potential energy. 

A simple example is illustrated in figure 84(A). A container is filled with 
liquid to a height h t . At the bottom of the container is a small hole. What is 
the velocity v 2 of the outflowing liquid ? If we construct a tube of flow as 
shown the conditions at the top of the tube are p t = atmospheric = A, 
height = Aj. velocity t>, ~ 0. At the bottom of the tube/> 2 = A, height = 0, 
velocity = v 2 . Starting with equation (12.6), 

-+ivl+h lg = ^ + ivl + h 2 g, 
P P 

-+0+h lg = -+\v\+Q, 
P P 

v\ = 2gh t . (12.7) 

The flow of an ideal liquid does not involve any particular molecular model; 
density is the only property concerned, and we shall not discuss it further. 
We may, however, remark that in general an increase in flow velocity is 

224 Gases, liquids and solids 

accompanied by a drop in pressure. This accounts for the suction experienced 
by a ship approaching a quay - the water flows faster at the regions of nearest 
approach. It also accounts for the downward curvature of the flight of a 
tennis ball that has been given top spin. Again, because the top surface of 
an aerofoil is longer, the air velocity in flight is higher and the pressure lower 
than over the bottom surface; this produces aerodynamic lift. 

12-2 Flow in real liquids, viscosity 

2-2-1 Steady and unsteady flow, turbulence 

Real liquids experience a viscous resistance to flow. We have already defined 
viscosity in discussing the behaviour of gases but we may recapitulate here. 
If the velocity gradient between two neighbouring planes is du/dz the force per 
sq. cm. to overcome viscous resistance is 

/«*£. (12.8) 

where i; is defined as the viscosity. Fluids which obey this relationship are 
known as Newtonian fluids. The dimensions of n are 

M = ML^T- 1 . (12.9) 

If the flow is steady and stable the work done is expended solely in over- 
coming the viscosity, and the work appears as heat. If, however, the velocity 
of flow is too high or there are other unfavourable features, a series of vortices 
may develop in the liquid. Some of the work is expended in providing kinetic 
energy for the vortices. Vortices often assemble on a solid boundary and this 
is known as the boundary layer. The condition for turbulent motion was first 
established by O. Reynolds. If a liquid of density p, viscosity q flows along a 
channel of lateral dimension a cm., there is a critical velocity at which orderly 
streamlined flow changes to turbulent motion. By dimensional analysis it is 
easy to show that this occurs when 

»< = -■ (12.10) 

The quantity k is called Reynold's number. It is dimensionless and is fairly 
constant over a wide range of conditions. Its value is of the order of 1000. 

For — < k the flow is streamlined. 

For — > k the flow is turbulent. 

Turbulence is essentially a condition of instability and the force involved in 
flow is not necessarily a clear indication of whether turbulence has begun or 

225 Liquids: their flow properties 

not. The force is determined primarily by viscous or inertia! factors. An inter- 
est ing simple example of these two conditions is furnished by considering the 
steady movement of a solid sphere of radius a through a fluid. The resistance 
to motion can be derived by dimensional arguments. One solution for the 
resistive force is 

F = Aanv. (12.11) 

This is Stokes' law; analysis shows that A has the value 6k. 
Another solution is 

F = Bv 2 pa 2 . (12.12) 

This does not involve viscosity but kinetic energy, the force being determined 
by the momentum of the oncoming fluid. 

12-2-2 Viscous flow through a tube 

Consider a cylindrical tube of length /, radius r. Liquid enters one end at 
pressure />, and leaves the other end at pressure p 2 . If the flow is uniform the 
streamlines are all parallel to the tube axis, there is no slip of liquid at the solid 
boundary and the velocity profile is parabolic. The rate of flow of liquid per 
sec. is 

Poiseuille first used this equation to determine the viscosity of blood in horses' 

12-2-3 Rotation of a shaft in a bearing: hydrodynamic lubrication 

Consider a shaft of radius a rotating with angular velocity o in a bearing of 
radius a+c [figure 85(a)] If the shaft carries a light load W and the space 
between shaft and bearing is filled with oil, they will remain concentric. The 
velocity gradient across the film is coa/c. The viscous force /per sq. cm. is this 
quantity multiplied by n. If the length of the shaft is /, the resultant tangential 
force that must be exerted at the edge of the shaft to maintain steady rotation is 

F = fx.2nal = —ti 2nal =—nalx 4k 1 , (12.14) 

c c 

where N = — is the number of revolutions per sec. 
In ^ 

We define the coefficient of friction jt as the ratio of F to W. Then 

M = — xn^-x4n*. (12.15) 

c W 

226 Gases, liquids and solids 






Figure 85. (a) Rotation of a shaft of radius a in a bearing of radius a+ c (Petroff). 
If the liquid has viscosity q and the angular velocity is (o the viscous force F that has 
to be overcome is proportional to ijm. (6) If the normal load is W the coefficient of 

friction is proportional to ~ or to — where N is the number of revolutions per 

second and P the normal pressure. At high pressures or low speeds the hydro- 
dynamic film is penetrated and the friction rises, (c) In practice a convergent oil 
wedge must be formed so that sufficient pressure can be developed in the liquid 
to support the applied load ; as a result the shaft is not quite concentric with the 
bearing (Reynolds). 

Since the nominal pressure P on the bearing is — we may write 


Ait 2 Nti 



For a bearing and shaft of fixed dimensions the quantity — is a constant so 


that ft is linearly proportional to Ntilp [see figure 85(6)]. 

This relation was first derived by A. A. Petroff in 1883 and describes ade- 
quately the behaviour of a lightly loaded bearing. However, it is evident that 
for loads which are not negligible, the journal cannot run in the central 
position. If the oil film is to support an appreciable load the journal must 
rotate eccentrically relative to the bearing, so that the lubricant is squeezed 
through the converging gap between the surfaces. This convergence in a 
viscous liquid, produces a hydrodynamic pressure in the oil wedge as O. 
Reynolds showed in 1886 and under suitable conditions it is sufficient to 
support the applied load.* 

* If the liquid were completely non-viscous the rotating journal would not drag any 
liquid with it. If we were to supply an external pressure and force the liquid through the 
gap. the convergence, as we saw in section 12.1. would actually produce a pressure drop 
in the convergent gap. It is primarily the viscous property of the liquid that is responsible 
for the development of a positive pressure in the liquid wedge. 

227 Liquids: their flow properties 

For a journal in a 'half bearing the pressure distribution is as shown in 
figure 85(r) and similar results are obtained for a 'full' bearing. It is seen that 
the (asymmetric) pressure distribution gives a resultant upthrust which would 
not occur if the journal rotated centrally in the bearing. Nevertheless the 
viscous resistance is not very different from that given by the simple PetrofF 
theory, and for a well-designed bearing friction coefficients of the order of 
0001 are obtained. 

A far more important difference between the Petroff and Reynolds treat- 
ment concerns the distance of nearest approach of journal to bearing. As the 
load increases the oil convergence must increase and this leads to a decrease 
in the separation. At some critical stage the separation necessary for ideal 
hydrodynamic lubrication may be less than the height of the surface rough- 
nesses. The hydrodynamic film is penetrated and contact occurs between the 
metals themselves or, at best, between metals covered by a thin adsorbed 
lubricant layer only one or two molecules thick. This is referred to as boundary 
lubrication and involves higher friction (/* = 0-1) and possible wear. By con- 
trast hydrodynamic lubrication gives a very low coefficient of friction 
(jt ~ 0-001 ) and, in principle, no wear at all. The region between the classical 
hydrodynamic and boundary regime is often referred to as the mixed regime. 
More recent work shows that in this region contact between asperities may be 
prevented by another mechanism. The very high pressures generated at the 
points of imminent contact can produce a prodigious local rise in viscosity. 
Consequently the lubricant cannot be extruded from precisely those areas 
where metallic contact might otherwise occur. In this way the range of success- 
ful lubrication is extended into a region of far more severe conditions. This 
type of lubrication, where high elastic stresses have a favourable effect on the 
viscosity of the lubricant, is known as elasto-hydrodynamic lubrication. When 
the elasto-hydrodynamic film itself is finally penetrated or destroyed we are 
back again at classical boundary lubrication where only molecular adsorbed 
layers remain. 

12-2-4 Simple ideas on the viscosity of liquids 

The viscosity of gases is explained in terms of molecular momentum-transfer 
from one plane to a neighbouring plane. The mechanical analogy is to con- 
sider two trains travelling parallel to one another at slightly different speeds. 
People are continuously jumping from one train to the other, so that those 
leaving the slower train, slow up the faster; whilst those jumping from the 
faster train speed up the slower. There is no net change in the numbers of 
people in any one train but there is clearly a momentum transfer tending to 
equalize the velocities of the two trains. Work must be done on the trains to 
maintain their difference in velocity. It is this process on a molecular scale 
which is responsible for the viscosity of gases. 

In liquids the momentum transfer process also takes place but it is a trivial 
part of the viscous resistance. The train model may still be used but now the 

228 Gases, liquids and solids 

jumping passengers find themselves caught by their coat-tails. They must grab 
hold of the other train and pull very hard before they can detach themselves 
from their own train. This will have a marked effect in tending to equalize 
the train velocities. In molecular terms one may say that in a liquid the mole- 
cules are so close together that considerable energy must be expended in 
dragging one molecular layer over its neighbour. No rigorous theory exists 
but a simple molecular model may be given. We first give it descriptively and 
in the next section deal with it analytically. 

Consider an instantaneous picture of the liquid. There is short-range order 
and we may represent a small part of two neighbouring molecular planes by 
an array similar to that of a solid (figure 86). To drag molecule A2 from its 


/ e \ 

A /' | \ 

© © © 



I— *- 


Figure 86. The-Evring molecular model of liquid viscosity (a) to transpose a mole- 
cule from position A2 to A2' intermolecular forces must be overcome. This involves 
a potential energy barrier e, (b) if a shear stress is applied to the right the potential 
energy curve is distorted, transitions to the right are facilitated, to the left hindered, 
(c) a shear stress / acting on a molecule occupying a nominal area a exerts a 
shear force fct, 

position to the neighbouring equilibrium site A2' involves overcoming the 
attractive forces of B2. There is thus a potential hill to be climbed, and energy 
is needed for this. (Note that in falling from the summit X to site A2' the 
energy that is restored appears as heat.) Occasionally molecules will have 
enough thermal energy to do this but there is just as much chance of the mole- 
cule jumping to site Al as to A2'. There is thus no net flow in any direction. 
If now we apply a shear stress to the right on the molecules in plane A 
mechanical work is done on the molecules. The thermal energy necessary to 
move the molecule to the right is reduced whilst that required to move it to 
the left is increased [figure 86(A)]. Thus thermally activated flow to the right 

Liquids: their flow properties 

is favoured. Another way of looking at the process is to say that before mole- 
cule A2 can move to site A2' it must make available a vacancy in the liquid 
to receive it. The energy to do this depends on the amount of opening up of 
the liquid that is necessary and on the latent heat of vaporization. Indeed the 
energy needed to make a vacancy available is roughly equal to -J to ^ of the 
latent heat of vaporization of the vacancy. 

12*2-5 The Eyring theory of liquid viscosity 

We consider a small cluster of molecules as shown in figure 86. In order to 
transpose molecule A2 to position A2' against the attractive forces of neigh- 
bours we have to surmount a potential energy barrier e. The rate at which 
this is possible as a result of thermal fluctuations is 

Cexp/--M, (12.17) 


where C is a frequency term. Eyring suggests that this is the vibrational 
frequency of the molecule and is given approximately by kTlh, where h is 
Planck's constant. Hence the frequency with which a molecule can jump to a 
neighbouring site as a result of thermal activation is 

^o = ^rexp-? (12.18) 

h kT 

When no stress is applied to the liquid there is an equal rate of jumping 
to left and right so that no net flow occurs. Suppose now a shear stress f is 
applied [see figure 86(c)]. If the average area occupied by a molecule is a the 
force on the molecule is /a. The only mechanical work we can perform is 
expended in carrying the molecule to the top of the potential barrier (on the 
other side of the barrier peak the molecule is assumed to give up all its energy 
as heat). The work done is 

f -f, (12.19) 

where X is the distance between neighbouring molecules in the plane. 

It follows that the thermal energy required for a jump to the right is reduced 

from e to e— - — , i.e., the rate of jumping is increased. Similarly the thermal 


energy required for a jump to the left is increased to e+ — so that the jump- 
ing rate is reduced. The net rate of jumping to the right is now, 


JT- kT 


230 Gases, liquids and solids 


For the exponentials in the square bracket we may take the first terms of the 
expansion. This gives 

We must now link Jf with the macroscopic flow process. The velocity 
gradient across the two molecular layers separated by a distance k x is 

, .^ ., velocity difference 
velocity gradient = 

(distance per jump) x (number of jumps per sec.) 


= K ' 

But by definition the viscosity tj is given by 

force per unit area , X t 
velocity gradient kJT 

Inserting the value of Jf from equation (12.22) we obtain 

Mi / e \ 


Putting ^1 ~ X and treating ai. x as the effective volume occupied by a molecule 
(»i»)» equation (12.25) becomes 

f -r.- p (£) 


where N is Avogadro's number, V m the volume per mole of the liquid 
and E the molar activation energy for surmounting the energy barrier. This 
relation is in full accord with the very marked decrease in viscosity observed 
with increasing temperature. 

The barrier energy E may also be regarded as the energy to create a hole 
in the liquid big enough to receive a molecule. This ought, therefore, to be 
comparable with the latent heat of vaporization {L)\ however, because there 
is already some free volume in the liquid, the work to open up a molecular 
hole is less than this. For a very large range of liquids 

E si (0-3 to 0-4)L. 
231 Liquids : their flow properties 

Consequently for a very wide range of liquids we may write 

hN /0-4£\ 
"=-^ eXP (^)- (12 - 27) 

However for liquid metals it is found that E is very much less than 0-4 L, 
presumably because the metal ions are much smaller than the metal atoms. 

12-2-6 Effect of external pressure on viscosity 

To examine the effect of pressure on liquid viscosity it is simplest to regard 
the energy barrier £ as the energy required to open up a hole for the trans- 
posed molecule. In the absence of applied pressure we may write equation 
(12.26) in the form 

tio = D exp ( — ) . (12.28) 


If now an external pressure P is applied the work to form a hole is increased 
by py h where V h is the volume of the hole. The thermal energy for activated 
flow is then increased from E to E+PV h - The viscosity then becomes 


" mDap (-7r) 


-tfoexpl-^J. (12.29) 

This type of relation is found to hold over a fairly wide range of pressures. 

If log e n is plotted against P a straight line of slope — is obtained. It turns 


out that V„ is about 1 5 per cent V m for simple liquids and about 5 per cent V m 
for liquid metals. Attempts to explain away these discrepancies are noi convin- 
cing. One must accept this as part of the inadequacy of the model. In broad 
outline it is surprisingly successful. 

The effect of high pressures in increasing viscosity can be very significant 
in lubrication. It is probable that many sliding mechanisms operate success- 
fully because local high pressures, which would normally be expected to 
squeeze out the lubricant and cause seizure, so increase the viscosity of the oil 
that it remains trapped between the surfaces. For example with heavily 
loaded gear teeth where elastic deformation in the contact zone may produce 
pressures of the order of 100,000 p.s.i., the viscosity of a typical lubricant 
may be increased a million-fold. In this way nature is much kinder to the 
engineer than he might have supposed. This type of elasto-hydrodynamic 
lubrication, (mentioned previously on page 228) is an area of great importance 
in current lubrication practice. 

232 Gases, liquids and solids 

12-2-7 Summary 

Although the model of liquid viscosity which we have discussed above is 
greatly simplified it brings out three important points. Liquid viscosity arises 
from intemnolecular forces and the ability of thermal fluctuations aided by an 
applied shear stress to produce flow in the direction of shear. Consequently 
the viscosity decreases if the temperature is increased. Secondly the viscosity 
increases if the pressure on the liquid is increased. This is because the molecules 
are pushed closer together, or, in terms of the alternative way of describing 
the molecular jumps, more work must be done to open up a vacant site. 

A third conclusion is that viscosity, being determined by intermolecular 
forces, is closely related to surface tension y and both are related to the latent 
heat of vapori2ation. For many simple organic liquids there is a fair cor- 
relation between log tj and y/Z where Z is a factor allowing for the size of the 

It may be mentioned that these models are most suitable for unassociated 
liquids with van der Waals bonding between the molecules. Highly polar 
molecules and long complicated molecules do not fit in so well. 

The basic differences between the mechanisms responsible for viscosity 
in gases and in liquids are clearly brought out by the following table. 

Table 26 Viscosity of gases and liquids 


Effect of temperature T 

Effect of pressure P 


ri increases as T* 



(/ decreases as 
log,»/ = A+- 

t] increases as 
log, n = A + kP 

12-3 Rigidity of liquids 

The molecules in a liquid jump about with a frequency of 10 10 to 10' 2 per sec. 
If the rate of shear is sufficiently great there may not be time for the molecules 
to advance to a neighbouring site. In this case the liquid will not show 
viscous flow, but will show a finite elastic rigidity. For simple liquids the rates 
of shear for this to occur are enormous but for liquids with large bulky 
molecules it may occur at moderate rates of shear. Silicone putties are a good 
example of this. There is a similar behaviour in polymers. The flow of poly- 
meric chains which occurs at slow rates of loading resembles the viscous flow 
of liquids: this may be impossible at high rates of strain The material will 
then deform elastically. Silicone putty for example will bounce with a very 

233 Liquids: their flow properties 

high resilience, since there is not enough time for the molecules to flow. If 
the material is in the form of a rod and is pulled rapidly it will first stretch and 
then snap in a brittle fashion. On the other hand at very low rates of deforma- 
tion, it will flow in a viscous manner. 

12-4 Non-Newtonian flow 

Many liquids encountered in daily life are non-Newtonian. This is particularly 
true of liquids with large bulky molecules or those which have some additional 
structure, e.g. colloids. A finite stress is often needed to break down the 
structure and initiate flow [see figure 87(a)]. Such materials are known as 



shear rata 

shear rate 

Figure 87. Schematic diagram of shear stress against shear-rate fcr (a) a Bingham 
fluid, (b) a Bingham fluid showing thixotropy. 

Bingham fluids. In many cases, once flow starts the molecules tend to orient 
along the direction of shear. The interaction of the molecules is reduced and 
the viscosity diminishes as the rate of shear increases [see figure 87(6)]. 
This is known as thixotropy and often occurs with paints. Brushing reduces the 
viscosity and the paint flows easily. On the other hand, the paint will not drip 
easily when left undisturbed. 

234 Gases, liquids and solids 

Chapter 13 

Dielectric properties of matter 

In this chapter we shall describe the main dielectric properties of matter 
and see how far they can be explained in terms of simple atomic models. 
As we shall see the main mechanism is the polarization of individual atoms 
and molecules by an electrostatic field. 

13-1 Basic dielectric relations 
3-1-1 Dielectric constant e and polarization P 

Suppose we consider a parallel plate condenser carrying a charge ■ 

unit area 
The field in free space between the plates [see figure 88(a)] is 

E, = Ana. (13.1) 

The potential difference between the plates is Anad and the capacity C per 
sq. cm. of condenser is 

c=J2!?*i = JL_ * (m) 

Potential Anad And 

We now consider what happens if we fill the intervening space with a di- 
electric. We find that the capacity of the condenser is increased by a factor s. 
This is known as the dielectric constant of the material. 

For an isolated condenser, since o remains unchanged, this means that 
the potential difference between the plates is reduced by a factor e, that is 
the field is reduced from E, = Ana to 

E= 4 ™. (13.3) 


This reduction in field is due to the polarization of the material. The field 
distorts the atoms and molecules, and orients existing dipoles in such a way 
as to create a field opposing the applied field. The resultant field is thus 
reduced.* We define a polarization P as the electrostatic moment per unit 

* If we had considered a condenser maintained in a circuit at a constant potential 
we could show that the insertion of the dielectric involves a further flow of charge such 
that the total charge, for the same potential, is increased by a factor e. Here again the 
free charges on the condenser plates are partly offset by the induced charges in the 

235 Dielectric properties of matter 

volume induced in the dielectric. As we shall see later P is very nearly pro- 
portional to the resultant held E. Consider a bar of dielectric of section A, 
length /. Its total moment due to polarization is 

Px volume = PAl =■ PAxl — end charge x length. 

Thus the moment per unit volume is equivalent to a surface charge P per unit 
area. We now see that the effective charge density on the plates is a— P 
[see figure 88(A)]. The resultant field is then 

£ = An(a-P) = *no-$nP. 






£, = 4no 

















T7^ + 









£„ = 

= 4n(a-P 


E~E< — 4nP 

Figure 88. (a) The field in an isolated condenser is E v . (b) When filled with a 
dielectric the material is polarized and the field is reduced to E . (c) Relation between 
E and £,. 

But from equation (133), eE = 4ror. Equation (13.4) becomes 
E = eE-AnP 



Thus a dielectric which reduces the electrostatic field by a factor e produces 


a polarization P which is times the real field. In what follows we shall 


show how P can be derived in terms of atomic models. Before doing so, 

however, we refer for completeness to the displacement vector D. 

236 Gases, liquids and solids 

M -2 Displacement vector D 

In passing from vacuum to a dielectric the field changes from E v to E. Taking 
the simplest case of a parallel block of dielectric normal to the field we see 
from figure 86(c) that 

E = E B -4nP, 

or E B = E+4nP = Ee. 

There is a discontinuity in the field. Another quantity called the displacement 
vector D may now be defined 

D = sE. (13.6) 

Then in vacuum, D v = «»E, = E v (since e, = 1), and in the dielectric 

D= eE= E„ (13.7) 

The quantity D is thus continuous. We shall not use it further. 

13-2 Polarization of gases 

3-2-1 Mechanisms of polarization 

We first consider the behaviour of gases since the molecules are, generally, 
so far apart that interactions between them may be neglected. Consequently 
each gas molecule 'sees' the dielectric field E. This polarizes the molecule in 
three ways: 

(i) By distorting the electronic cloud around the nucleus or nuclei of 

each molecule. We call the polarization due to this mechanism P e . 
(ii) By stretching or bending bonds in polar molecules; we call this P a 

Usually this is small (P„ ~ (MP.) and can be neglected, 
(iii) By orienting existing molecular dipoles: we call this P t . 

We first consider the electronic polarization P t . Suppose the molecule is 
initially non-polar. As a result of electronic displacement the individual 
molecule (or atom) will acquire a dipole moment p. As we shall see immediately 
p is proportional to E, so we may write 

p = «E, (13.8) 

where a is the molecular (or atomic) polarizability. If there are n molecules 
per cm. 3 , 

P t = np=> mE. (13.9) 


We at once note that the dimensions of P, are whilst the dimensions 


of E are , so they are dimensionally identical. Hence «a has zero 

(distance) 2 

237 Dielectric properties of matter 

dimensions or 
[«] = 

= IL? 

In fact as we shall see in what follows a is equal to (molecular radius) 3 . 

13-2-2 The polarization of neutral monatomic molecules 

Consider a neutral atom such as argon or neon, which consists of a central 
positive charge Ze and an electron cloud of charge -Ze [see figure 89(a)]. 
We assume that the electron cloud has uniform charge density p and extends 
over a sphere of radius r. If a field £ is applied the electron cloud experiences 
a force E Ze in one direction and the nucleus an equal force in the opposite 
direction. The charges are displaced relative to one another until the attractive 
force between them exactly equals the force exerted by the field. Let the 
displacement be x [see figure 89(b)]. The attraction between the nucleus and 




Figure 89. An atom of atomic number Z considered as a central positive charge 
surrounded by a negative charge of uniform density. The application of a field £ 
displaces the positive relative to the negative charge. 

the electron cloud is the force between Ze on the nucleus and that part of the 
electronic charge which is within the sphere of radius x. 

The latter is Ze I - J . The attractive force [see figure 89(c)] is then 
(positive charge) x (negative charge) 

/ = 

f=Zex.Ze (- 

1 ZV* 

ri x" 


238 Gases, liquids and solids 

This is balanced by the distorting force 

f=EZe. (13.11) 

Equating (13.10) and (13.11) we have 

E^~. (13.12) 

r 3 

But Zex is the charge times the distance between the charges, i.e., it is the 
atomic dipole p. Hence 

p = r 3 E. (13.13) 

A similar result is obtained if the atom is assumed to be a perfectly conducting 
sphere of radius r. Hence 

P. = nr 3 E. 

Combining this with equation (13.5) we have 

ox s- 1 = 4rmr 3 . (13.14) 

13-2-3 Dielectric properties and van der Wools critical data 

We now make a wide-ranging and rather unexpected correlation between 
s and critical gas data. We note that 4nnr 3 is three times the volume of the 
molecules in I cm. 3 of gas. The critical volume of the gas is 12 times the 

volume of the molecules. It follows that 4nnr 3 is equal to where p is the 

4p c 

density of the gas and p c the density at the critical point. Hence 

(«- 1) = J - . (13.15) 

According to Maxwell's theory of electromagnetic waves, at the same 
wavelength, the refractive index 0} is related to the dielectric constant e and the 
magnetic permeability p by the relation 

3H= v(e//). (13.16) 

For dielectrics p is equal to unity to within one part in 10*. Since for gases e 
is only slightly greater than unity we may write 

® = ve = v[l+(fi-l)] = l+i(«-l), (13.17) 

so that (3?-l) = i(fi-J). (13.18) 

239 Dielectric properties of matter 

Combining with (13.15) we have 



These quantities have been compared in Table 27 for inert (monatomic) 
gases, non-polar polyatomic gases, and for molecules with dipoles. The 
dielectric constant is found from static or low-frequency measurements. 

Table 27 Dielectric properties of gases 










pxlO 3 
g. cmr 3 


g. cmr 3 

1 P 

4p c 

xlO 4 

2(M- 1) 
xlO 4 



























H 2 







N 2 






o 2 






Cl 2 






co 2 







NH 3 






with dipoles 

so 2 






H 2 






There are three clear conclusions. First, for the inert and non-polar 

molecules the agreement between and the observed values of (e- 1), see 


columns 5 and 6, is reasonably good except for He where the inner two 

electrons are very tightly bound. We conclude that our calculation of the 

molecular polarization is valid for polyatomic molecules as well as for 

240 Gases, liquids and solids 

monatomic molecules. Secondly, for all these gases there is very good 
agreement between (e- 1) and 2($?- 1) - columns 6 and 7. Since the optical 
refractive index is a measure of the interaction of the molecules with the high- 
frequency electric field associated with light waves we deduce that this is the 
same as the electronic polarization produced by a static or low-frequency 
electric field in these cases. 

Thirdly we note that polar molecules give good agreement between 

4 A: 

and 2(01 - 1) - columns 5 and 7 - since both are a measure of high-frequency 
electronic polarization. There is poor agreement between (e— 1) and 2(0t— 1) 
- columns 6 and 7 - since e is a low-frequency determination and includes the 
orientation effects of the molecule. By contrast 01 involves optical frequencies 
where only the electronic polarizations can respond. Indeed the difference 
between (e— 1) and 2(0t — 1) for ammonia suggests that the orientation ef the 
dipole contributes 63 x 10"* e.s.u. to the low-frequency dielectric constant. 

13-3 Polarization of polar molecules 

3-3-1 Apart from the electronic polarization, polar molecules show two other 

(i) The stretching or bending of bonds. This produces a polarization P„ 
which is generally very small. We shall not attempt to estimate it 

(ii) An orientation of existing dipoles. The polarization P t due to this is 
large and may be as much as 100 P e . 

Materials in which existing dipoles are oriented by the applied field are 
sometimes referred to as para-electric materials. In what follows we provide 
a simplified model by means of which we can estimate the magnitude of Pt 
The simplest example of a dipole (see figure 90) is a diatomic molecule 

—e +e 

Q .,,., Q 

Figure 90. A simple electrostatic dipole. 

with a charge e at one end and —e at the other and a separation of say 1 A. 
The permanent dipole has a moment 

p = charge x separation 

= 4-8xlO- 10 xlO-» 

= 4-8xl0-**e.s.u. (13.20) 

241 Dielectric properties of matter 

In the absence of a field these dipoles are randomly oriented and show no 
resultant moment. If an electric field could orient all the molecules we 
should obtain a polarization per cm. 3 given by 

P d = tip. 
For a gas at s.t.p. n is about 3 x 10 19 . Hence 

P„ =i 140 e.s.u. 
The field E„ produced in the dielectric by these dipoles is 

E„ - AnPt = 1700 e.s.u. = 5xl0 5 volt cm." 


This would be more than enough to ionize most gases, but experiments show 
that it is virtually impossible to ionize a gas in this way. The reason is that the 
calculation ignores thermal motion which vigorously opposes the lining-up 
of the dipoles. In most practical situations the polarization achieved is very 
much less than that calculated above. 

13-3-2 Orientation of dipoles in an electric field: the Langevin function (para-electric 

We consider the effect of a field E on an assembly of dipoles of moment p. 

The field attempts to orient the dipoles, thermal motion to maintain disorder. 

We first calculate the energy of a single dipole in the field if its axis makes 

an angle 6 to the direction of E [see figure 91(a)]. Let the dipole consist of 





Figure 91 . The orientation of a dipole of moment laq in a field of strength E. 

charges q and -a separated by a distance 2a. Then p - laq. The potential 
energy is defined as work done on the system by external forces. The forces on 
the charges are qE and -qE [figure 91(6)]. If both charges were at some arbi- 
trary position, say at O, we should do work -qEx on the positive charge and 
qE(-x) on the negative charge in moving them to their present positions. 

242 Gases, liquids and solids 

The total potential energy e is 

e = -2qEx = -2qEaCO&6 
or e=-£pcos0. (13.22) 

We now consider how much space is occupied by orientations between and 
0+d0. Construct a sphere of arbitrary radius R [figure 91(c)]. The solid angle 
subtended between and Q+dO is 

area of annulus 2nR sin RdO 

dco = : = -, 

R 1 R 1 

or dco = 2n sin d0. (13.23) 

We now apply the Boltzmann distribution. The number of dipoles with axes 
between 6 and 0+dO is 

dn = B exp ( — -I x (element of space) 

= B exp I - 1 2* sin d0. (13.24) 

As this is a rather clumsy expression we make the following substitutions: 

x = P -* I 

kT k (13.25) 

z = cos so that <fe = - sin d0 J 

Then equation (13.24) becomes 

dn = -2nBexp(zx)dz. (13.26) 

We find B by integrating rfw for from to ?r, that is for z from +1 to — 1. 
This gives n, the total number of dipoles per cm. 3 

n = 2nB \ exp (zx) dz. (13.27) 


(The change in sign for n follows because we have reversed the order of the 
limits of integration.) Because of symmetry the resultant polarization normal 
to E will be zero, and only the polarization parallel to E will be additive. 
Each dipole contributes a dipole moment parallel to E of amount p cos 0. 
Thus the resultant dipole moment per unit volume P t will be 


Pa = P cos 8 dn 


= 2nBp I z exp (zx) dz. (13.28) 

243 Dielectric properties of matter 

We can express the ratio of P t to n 

Pi _ equation (13.28) _ pfze'Vz 
n equation (13.27) Je"dz ' 

where the integral is from z = — 1 to z = + 1. 
In most practical situations zx is small so that 
e" ~ 1+zx. 
Equation (13.29) then becomes 

Ft = V pjz(l+zx)dz l +1 _ x 
n [ ](l + zx)dz J_i P 3' 

Substituting from equation (13.25) for x we have 
np 2 E 

P* = 






We may at once note that for most dielectrics E cannot exceed about 
10 5 volt cm. -1 Oi 300e.s.u., otherwise breakdown occurs. For a typical 
value of p = 4-8xl0- 18 e.s.u., using k = 1-4X10" 16 , and T= 300°K., 
we have 


si 003, 

so that our assumption implied in equation (13.30) is valid. 
Exact integration of equation (13.29) gives the Langevin function 

t *-H£)-sl- 



10 kT 

Figure 92. The Langevin relation for the polarization of dipoles produced by 
a field E. 

244 Gases, liquids and solids 

This function is plotted in figure 92 and shows that under normal conditions 
the linear part of the curve, far removed from saturation, is the only part 
that is observable. The contribution of P d to the dielectric constant may now 
be calculated. We have 

P ^ ( s -»< E= n P 2E 

' An 3kT 

or <>-».-*£. (13-34) 

For ammonia at s.t.p. we have n = 3 x 10 19 , p si l-5x 10~ 18 e.s.u. Then 

(e-l) d = 67xl0-*e.s.u. 

This agrees extremely well with the value quoted in the discussion on page 
241 following Table 27. Measurements of the dielectric constant are often 
used by the chemist to deduce the value of the molecular dipole. 

The orientation polarization can, of course, be increased without exceeding 
the breakdown of the dielectric by reducing the temperature. This cannot, 
however, be carried very far since most polar substances solidify fairly readily 
and under these conditions molecular orientation becomes very difficult or 

Effect of frequency of electric field 

We can at once see the powerful effect of frequency on the polarization. 
We write 

P = P t +P.+P t 

= f£ B+ P. + m»E. 

In a static or low-frequency field all these processes will exert their full part. 
However, the orientation of the whole molecule is relatively slow, and as the 
frequency increases the orientation lags behind. When the frequency reaches 
a value of about 10 12 the dipoles are unable to follow the oscillations of the 
field : only random orientations remain and these contribute nothing to the 
resultant polarization. Of the polarization, P, only P a and P e remain. At a 
somewhat higher frequency the stretching of the bonds becomes too sluggish 
and P a drops out. Only P e remains above a frequency of 10 15 . This is in the 
optical range. The behaviour is sketched in figure 93. However, this ignores the 
fact that both P„ and P e may show resonance effects. In the next section we 
shall consider only P e and show that resonance occurs at a critical frequency 
and that this accounts for optical dispersion and anomalous dispersion. 

245 Dielectric properties of matter 

Prf + p. + p. 





Figure 93. Effect of frequency on polarization ignoring the effects of resonance. 

13 -4 Optical dispersion and anomalous dispersion 

We now analyse the effect of an alternating electric field on the electronic 
polarization P e ; the treatment is similar to that given by G. Joos. For 
simplicity we consider a single electron, mass m, charge — e, bound to its 
equilibrium position relative to its nucleus by a force of force constant k. 
In free oscillation the equation of motion of the electron is 

mx+kx = 0, 

or x+a>lx = 0, (13.35) 

where co is the natural angular frequency (radians sec. -1 ) and col = k/m. 
The natural frequency v is given by 

Vo = — cycles sec. -1 (Hertz). 


We apply a sinusoidal electric field 
E = E sin tor, frequency v i 

Ignoring any 'fMctional' effects the equation of motion is 

, — e 
X+copc =b — E sin cot. 






246 Gases, liquids and solids 

The steady-state solution is 

— eE sin cot I 

x = 


a>l~(0 2 ' 

The dipole moment per atom is simply 

p = —ex. 

P - 

e 2 E sin cot 

m(col—o} 2 ) 4n 2 nKyl— v 2 ) 




l/ — VJ I -Til ntyr - 

The volume polarization P ( is simply n times this. Using the relationship 

(*-l) = 2(^-l) = 4*J=^, 
t, Jo 

and substituting for /> from equation (13.41) we obtain 


?=1 + - 



This is plotted in figure 94(a). It is seen that approaching v the refractive 







Figure 94. (a) For an undamped system, resonance at frequency v produces a 
change in 0t from +oo to — co. (b) With a damped system the transition is sharp 
but the infinite values do not occur. 

index rises to plus infinity and just beyond v it falls to minus infinity. This 
is the absorption frequency of the atom and is generally of the order of 10 15 . 
In practice radiation resulting from the oscillation involves a certain amount 

247 Dielectric properties of matter 

of damping or 'friction', and this prevents infinite values of ^ from being 
reached. The main features are now shown in figure 94(6). The refractive 
index increases with frequency over the region AB; this is the region of 
'normal' dispersion. The sudden change in behaviour along CD was, at one 
stage, considered anomalous and was so called. The above analysis shows, 
however, that 'anomalous' dispersion is just as normal as 'normal' dispersion. 

The frequency v at which electronic resonance occurs is the same as that 
involved in the derivation of van der Waals forces. For this reason these 
forces are sometimes called 'dispersion' forces. 

Similarly the frequency at which a resonance effect occurs in P a is of the 
order of 10 13 , so dispersion also occurs in the infrared region. The combined 
behaviour for a dipolar molecule with a single electronic and atomic 
absorption line is shown schematically in figure 95. 


P.+ P. 

in P. 


in P. 



10 12 c/s. 


~10 15 c/s. 


Figure 95. Effect of frequency on polarization of a dielectric allowing for resonance 

13-5 Dielectric properties of liquids and solids 

13-5-1 Weak interaction 

We now consider the dielectric behaviour of matter in the condensed state. 
We first treat the simplest case in which non-polar molecules are involved 
so that interaction between neighbouring atoms or molecules is weak. In 
this case we consider only the electronic polarization per atom 

P = r*E, 


248 Gases, liquids and solids 

where r is the radius of the individual atom. For simplicity, to obtain an order 
of magnitude value for the number of atoms per c.c, n, assume a simple 
cubic structure. 

' \3 



The polarization per unit volume, using equations (13.43) and (13.44), is 

P = i«»=f. (13.45) 



e-1 = = -. (13.46) 

E 2 


e = 1 + - =i 2-6, 

M= Ve s 1-6. (13.47) 

This analysis shows that for many simple liquids and solids the dielectric 
constant will be of order 2-3 and the refractive index 1 -4 to 1 -7. Clearly in a 
crystal which has well defined structural anisotropy, so that the polarizability 
depends on crystal direction, there will be a corresponding dependence of e 
and &t on crystal direction. This is the basic cause of birefringence. 

Strong interaction: the Clausius-Mosotti equation 

When atoms or molecules are far apart as in a gas, they 'see' the average 
dielectric field E. When, however, they are close together as in a liquid or 
solid, the atomic (or molecular) dipoles may themselves interact. For example 
if polarization produces a local dipole of strength p = r 3 E, the field produced 
by this dipole at a distance r is of order of magnitude p/r 3 , i.e., about E; that 
is, the field produced at a neighbour by an induced dipole may be about as 
large as the field itself. This increases the polarization effect. 

We have now to consider how to estimate the resultant field acting on 
each molecule. The treatment is due to Lorentz. We choose a molecule O and 
construct a sphere of radius R large compared with the molecular separation 
(see figure 96). The resultant effect at O is the sum of (a) the field produced by 
the material within R and (b) the field produced by the material outside R. 
The latter can be treated as a continuum since it is so far away from O. The 
field due to this is the same as the equivalent surface charges over the spherical 
surface. The calculation is as follows. 

249 Dielectric properties of matter 

resultant £ 


0+ — P 





"I dielectric 

Figure 96. Charge distribution in a spherical cavity within a dielectric. 

p + 

— a 














Construct an annulus on the surface of the cavity making an angle 0, 
0+d9 with the direction of the polarization field. The surface charge density is 

The charge on the annulus is 

-P cos 2nR sin RdO. 



This produces a coulomb field at O pointing along OA. The vertical com- 
ponents cancel out, the horizontal ones are additive. The horizontal com- 
ponent at O, for the annulus, is 

Pcos BlnRsm 0Rd9— cos 
R 2 

= 2nPcos 2 0sin0d0. 


For the whole sphere, for ranging from to n, the net field E' at O due to 
the surface charge is 


E'=2nP\ cos 2 0sin0<W = 2jiP|— ~^-^\ =^. (13.51) 

r-cos 3 0"i* 

The resultant field at O is thus E+E' or 

We now have to consider the contribution (a) due to material within the 

£+4 f- 

260 Gases, liquids and solids 

sphere. Lorentz showed that for a cubic array of dipoles or for a random 
array of dipoles as in a liquid or a glass, the local field is zero. For such 
materials we may, therefore, write 

The induced dipole per molecule is 

P = «E lmU (1334) 

/. P = np = na U+^f] • (1335) 

Hence - = —^— . (1336) 

The basic relation between P and the average field 2? in the dielectric is 

. AnP Anna n-xvn 

(e-1) = — = — -r . (1337) 

E 1— \nna 




Hence (e+2) = (<s-l)+3 = - — 2 . 

1— fwta 

Taking the ratio of equations (1337) and (1338), we have finally 

«— 1 _ Anna 
e+2 3~" 

This relation is fairly well obeyed for polar liquids and for dilute solutions 
of dipolar molecules in a non-polar solvent, as well as for ionic solids of cubic 
structure. It is known as the Clausius-Mosotti equation. 

Before we leave this section we may form some idea of the importance of 
interaction compared with the results obtained if interaction is ignored. In 
the latter case, as we saw above, 

e- 1 = An - = Ann*. (13.60) 


Consider three cases: 

(a) Weak interaction, say e a 1; we see that equation (13.60) is identical 
with equation (1339). 

(b) Slight interaction, say e = 3; equation (1339) then gives a value for 
(e- 1), which is f times bigger than that given by equation (13.60). 

(c) Large dipole interaction, say e = 70; equation (13.59) gives a value for 
(e- 1) which is 24 times larger than that given by equation (13.60). 

251 Dielectric properties of matter 

There is yet a further type of interaction which we have not discussed here, 
but shall refer to briefly in the next chapter. This is the analogue of ferro- 
magnetism where interaction is intrinsically so large that even without the 
presence of an applied field the material is strongly polarized. It would 
correspond to a situation [see equation (13.58)] where \nno. m 1 so that 
e — >■ oo. Such materials are known as ferro-electrics. A ferro-electric material 
such as barium titanate has a dielectric constant of over 1000. 

13-5-3 Effect of frequency 

With matter in the condensed state frequency has the same effect on the 
dielectric constant as it has on gases. However the dipole polarization in 
liquids is usually more sluggish than in gases - with solids it is often completely 
absent since rotation of the whole molecular unit within the lattice is difficult 
or impossible particularly at temperatures well below the melting point. 
Results for water and ice are shown in figure 97. 



water at 20 °C. "X 





\ice at - 












I • 1 .. 

10 3 

10 s 

10 9 


Figure 97. Variation of dielectric constant with frequency, for water at 20°C- 
and for ice at — 5°C. and — 40°C. The large dielectric constant at low frequences 
at —40° C. shows that molecular rotation is still possible (see p. 178). 

13 -6 Summary 

We may summarize the ideas described in this chapter in the following way. 

1. The dielectric properties of matter arise from the polarization produced by 
an electrostatic field. Polarization occurs in three ways: 

252 Gases, liquids and solids 

(a) Distortion of electron clouds; all atoms (or molecules) show this effect. 

(b) With polar molecules stretching or bending of bonds can produce a small 
contribution to the polarization. 

(c) With polar molecules orientation of the molecule may produce a very large 
polarization: the field has to compete with thermal randomization. 

2. Temperature has no effect on (a) since this involves electronic properties. 
Increasing temperature resists the polarizing influence of the field for both 
(b) and (c). It is particularly marked with (c). However, the molecular orienta- 
tion cannot be increased indefinitely by lowering the temperature since 
orientation of the molecular dipoles may become impossible when the 
material is solid. 

3. Frequency of the electrostatic field has a marked effect on all types of 
polarization. Orientation becomes increasingly difficult as the frequency is 
raised and becomes impossible at frequencies above about 10 12 cycles 
sec. -1 The stretching or bending of bonds also becomes unable to follow the 
field at frequencies above about 10 14 cycles sec. -1 , and only the electronic 
polarization remains up to a frequency of 10 15 -10 16 cycles per second. This 
is in the optical range and is therefore the only part of the dielectric mechanism 
which plays a part in the refractive index of the material. There is usually a 
resonance effect at frequencies in this region and this accounts for optical 
dispersion. The bond stretching (or bending) mechanism also shows a reson- 
ance effect but this generally occurs in the infrared. 

4. With matter in the condensed phase the behaviour is complicated by 
neighbour-neighbour interaction. This can greatly increase the effective 
polarization of the material. 

253 Dielectric properties of matter 

Chapter 14 

Magnetic properties of matter 

Magnetic properties are more difficult to interpret than dielectric properties. 
This is partly because the basic mechanisms are more involved and partly 
because the historic development of the subject has led to confusion, particu- 
larly between the field H and the magnetic induction B. Fortunately the 
difference between H and B is not of great importance in explaining the 
mechanism of diamagnetism and paramagnetism. Nevertheless we shall 
attempt to keep the issue clear. 

14- 1 The magnetic equations 

14 • 1 • 1 Field, permeability, magnetic induction and susceptibility 

We begin with the concept of a magnetic field, which should be denned in 
terras of the couple exerted by the field on a small loop carrying a specified 
current. This is not very practical and it is more convenient to think in terms 
of the force exerted on a magnetic pole. Isolated poles do not exist but long 
thin magnets with spherical end-pieces behave as though the poles were 
isolated at each end. Such magnets may be used to define a unit magnetic 
pole: two unit poles, one centimetre apart in air, exert a force on one another 
of 1 dyne. We can then define a magnetic field H in air as the field which 
exerts a force mH on a pole of strength m. We may wind a coil and find that 
for a specified current it exerts a force mH on the same pole in air. The field 
due to the current is then defined as H. 

If in this coil we insert a piece of magnetic material the field is increased. 
The pole of strength m, if placed near the magnetic material, will experience 
a force greater than mH. We may approach this result from a different angle. 
Suppose the coil originally possessed self-inductance L. This arises from the 
change in magnetic flux cutting the turns of the coil when the current is 
changed. Subsidiary experiments with the coil in air show that the flux 
is proportional to the magnetic field H which it produces. Suppose now the 
whole of the space in the coil is filled with magnetic material. It is then found 
that the self-inductance is increased. Suppose the factor of increase is n; then 
/i is defined as the magnetic permeability of the core material. Clearly the 
magnetic flux has been increased by this factor. We now invent a new quantity 
B known as the magnetic induction and say that, in the magnetic core, 

B = »H. (M.1) 

254 Gases, liquids and solids 

It may be shown that B is continuous across a normal interface exactly as 
for D. The force exerted on a unit pole outside the core will be increased but 
this increase will depend on the shape of the end-piece of the core. 

The clearest and least problematical situation is that in which the coil is in 
the form of a torus (see figure 98). If the space is filled with magnetic material 

Figure 98. Effect of magnetic field on magnetic material. 

there are no free ends. The increase in magnetic flux is due to polarization 
of the material imparting to it magnetic moment per unit volume M. The 
effect of this on the magnetic flux may be considered by cutting a very thin 
slice out of the core normal to the direction of the field. In this gap the 
original field H is augmented by M. This is equivalent to a distribution of 
magnetic poles over the free surface of amount M per sq. cm. (in this case it is 
a fictitious equivalence since free poles do not exist). The additional field 
produced is AnM so in the gap the field is H+4nM. Since B is continuous 
across the interface and n in free space is unity, 

B = H+4tiM. 


This is closely analogous to the dielectric relation given in equation (13.7), 

D = E+4nP, 

and the similarity has been generally stressed. It is, however, misleading for the 
following reasons. First, consider the forces experienced by a charge or a 
magnetic pole within a medium. Within a dielectric the forces on a charge 
are reduced; within a magnetic material the forces on a pole are increased. 
This is because the electrostatic forces are determined by (the local values of) 
E and magnetic forces by B. The latter point has been proved in a direct way 
by F. Rasetti in 1944 in his study of the deflection of cosmic rays through 
magnetized iron (F. Rasetti, Physics Review, vol. 66, 1944, p. 1). For fast rays, 
where the cosmic particle sees the material as a continuum, the deflection 
is found to correspond to B and not to H. This is consistent with the view 
that magnetized particles behave like circulating currents (see below). 

265 Magnetic properties of matter 

A second difference concerns the direction of the field. If electric charges 
are placed on a dielectric the direction of the electrostatic field is always from 
positive to negative whether it is observed inside or outside the dielectric. 
With a permanent magnet, however, the situation is different: outside the 
material the field is in the direction north to south; within the material the 
inductive field (the magnetic induction) is from south to north and a unit 
north pole would experience a force in that direction. Here again the primary 
magnetic elements behave like circulating currents and the permanent magnet 
as a whole behaves like a solenoid. 

Before returning to the magnetic equations we may note the following 
further differences between electrostatic and electromagnetic concepts: 

(a) there is no such thing as an isolated magnetic pole 

(6) there is no magnetic analogue of a condenser 

(c) there is no electrostatic analogue of a solenoid. 

Although B is of such primary importance we still need to be able to 
express magnetic phenomena in terms of H. This is largely because most 
magnetic experiments involve the use of solenoids where H is known in terms 
of the number of turns and the current. We may eliminate B from equation 
(14.2) by writing 

B- /iH= H+4nM 
so that 

0i-1W=4tzM. (143) 

Since the magnetization M is a function of the field H we may write 

Af = X H (14.4) 

where x is termed the magnetic susceptibility. Then from equations (14.3) 
and (14.4) we have 

H-\=4n— = Anx- (14.5) 


Materials may be divided into three main classes according to their values 
of x. 

(a) Diamagnetic. % is small and negative. The material is weakly repelled 
by a strong magnetic field. All substances possess a diamagnetic 

(6) Paramagnetic, x is small and positive. Such materials are weakly 
attracted by strong magnetic fields. 

(c) Ferromagnetic, x is large and positive. The material is strongly 
attracted by a magnetic field. 

266 Gases, liquids and solids 

In practice x is roughly constant only for feebly magnetic materials, i.e., 
for diamagnetic and paramagnetic substances. For ferromagnetic materials x 
varies very markedly with the field itself. In what follows we shall derive 
expressions for x in terms of atomic models. 

Force exerted on a current by afield 

A current i is defined as the charge per sec. passing through a conductor. 
If a short length dl of such a conductor is placed in a field of induction B 
at an angle it is found experimentally that it experiences a force 

Bisinedl, (14.6) 

in a direction at right angles to the plane containing B and dl [figure 99(a)]. 



*.Bi smBdl 

Figure 99. (a) A field B acting on a current element / produces a force normal to 
the plane containing B and /. (b) Calculation of moment exerted by a field B on a 
coil carrying current /. 

The force is determined by B not by H. Consider a coil of irregular shape in 
the plane xy, the field B pointing in the x direction [figure 99(Z>)]. There is a 
force on the element P of amount 

iB sin Odl= iBdy, 


out of the paper and an equal force at Q into the paper. These two forces 
exert a couple 

iBdyXi = iBdA, 


where x% is the distance PQ and dA is the area of the strip between P and Q. 
For the whole coil the couple is then BiA where A is the area of the coil. Thus 
the coil behaves as though it has a magnetic moment iA so that in a field of 

257 Magnetic properties of matter 

induction B it experiences a couple BiA . (This may be extended to a coil which 
is not planar and leads to the more general concept of the equivalent magnetic 
shell. As we do not need this in what follows we shall not pursue it further.) 
Ever since A. M. Ampere, it has been considered that the magnetism of atoms 
is due to circulating currents within the atom; consequently every north pole 
has associated with it an equal south pole. For this reason it is not possible to 
envisage the existence of isolated magnetic poles in the way that there are 
isolated charges in electrostatics. The earlier view regarded the circulating 
currents as arising from electrons travelling in orbits; later work has included 
the electron spin which, from this point of view, can also be regarded as a 
minute circulating current. 

It follows from this discussion that the force on a magnetic dipole depends 
on B. When, however, we deal with atomic dipoles we must consider the true 
or local value of B at the atom itself. For paramagnetic and diamagnetic 
materials, as we shall see, this raises no real problem since n is extremely 
small and B and H are almost equal. 

14-2 Diamagnetism: Langevin's treatment 

All atoms possess a diamagnetic component. This is because an applied field 
always modifies the circulating 'electronic' current in such a way that it 
opposes the field. We consider the simplest case, an electron of charge e 
e.s.u., moving with velocity v in a circular orbit of radius r. If t is the time for 
the electron to describe one revolution, e behaves like a current of magnitude 

e \ ev I .. . „ v 

i=-x- = — x-, 04.9) 

T c 2nr c 

where c is the velocity of light and is introduced to express the current in 
e.m. units. 
The magnetic moment of the orbiting electron is 

p = nrH - ^ . (14.10) 


Consider now the effect of a field of induction B which is normal to the orbit. 
The flux 4 in the circuit is 

* = nr*B. (14.11) 

If the field varies with time an e.m.f. will be induced in the circuit of 

e.m.f. = = . (14.12) 

c dt c dt 

This e.m.f. acts around the current loop and is equivalent to an electro- 
258 Gases, liquids and solids 

static field E, where 

e.m.f. = 2nrE 


~ c dt ' 


r 1 dB 
E= --X-X-— . 
2 c dt 



This field exerts a force eE on the electron and this in turn produces a change 
do in its velocity: 

„ do re dB 

eE = m — ss x — , 

dt 2c dt 


re dB 

x — . 

2mc dt 




Consequently in the time that B changes by AB, v changes by Av where 

re . _ 

A» = AB. 


According to equation (14.10) a change in v produces a corresponding 
change in the magnetic moment of the orbit: 

Ap = — Av. 

Combining with equation (14.17), 



Ap = AB. 



We see that Ap is always in a sense which opposes AB. Consider for example 
two identical orbits with opposite orbital momenta. Each orbit has a magnetic 




resultant =0 

+ P+^P 




Figure 1 00. Sketch showing nature of diamagnetism ; (a) when no field is applied 
the net moment is zero, (b) when field AB is applied there is an additive change in 
the magnetic moments of both orbits. 

258 Magnetic properties of matter 

moment p but the net moment is zero. If a field AB is applied one orbit is 

slowed down and the other is speeded up: the net change is additive for the 

two orbits (see figure 100). 

There is another point of interest. If the orbital velocity is reduced by At>, 

. .- , (mv 2 \ 2mvAv. 
the centrifugal force on the electron is reduced by Al I = From 


equation (14.17) this equals . But the field AS itself exerts a radial force 



on the electron of amount: velocity x charge x AB = . This exactly 


compensates for the reduced centrifugal force: consequently if the initial 

orbit is stable under some specified force between nucleus and electron the 

diamagnetic mechanism does not introduce any change. 

For diamagnetic materials B 2; H to 1 part in 10*. Equation (14.18) then 


Ap=-^—AH. (14.19) 

4mc 2 

By definition the magnetic susceptibility per electron is simply . If there 


are n atoms per unit volume the susceptibility per unit volume is 

Ap __ n , 

where the summation is for all the orbital electrons in each atom. 

If the orbits are not normal to B each orbit undergoes precession and x is 
slightly decreased. Similarly if the orbits are not circular a lower value of x is 
obtained. A better value for x is 

6 z_l mc 2 

We may at once derive an order-of-magnitude value for x- If a gram-atom 
occupies 10 c.c, n = 6 x 10". If there are say 10 orbital electrons and r is of 
order 10 ~ 8 cm. we have 

X= -(2to3)xl0" 6 . (14.22) 

This is a typical value for most solids and liquids. In the gaseous state the main 
change is in n. At s.t.p. n is about 10 3 times smaller than in the condensed 
state so that for gases at s.t.p. 

X =: 10-». (14.23) 

Note that x is independent of temperature since the internal structure of the 
atom is virtually unchanged by normal temperatures. 

In the model described above, diamagnetism is attributed to the influence 
of the magnetic field on the orbital momentum of the electrons. What then 

x = n ^- = -ZJ"LL- t (14.20) 

260 Gases, liquids and solids 

happens to the j-electrons, which possess zero orbital momentum? A detailed 
quantum theory treatment shows that ^-electrons possess zero angular 
momentum only in the absence of a magnetic field. Upon the application of a 
magnetic field they acquire a small amount of orbital momentum. This pro- 
duces an equivalent amount of diamagnetism to that calculated above. Thus 
it follows that all the electrons contribute to diamagnetism. 

14-3 Paramagnetism: the Langerin function 

If an atom possesses a permanent magnetic dipole because the electrons have 
either orbital momentum or spin, the dipole will orient in an applied magnetic 
field and produce a magnetization in the same direction as the applied field. 
Materials which are made of such atoms are attracted (weakly) by a magnetic 
field and are known as paramagnetic substances. As in the case of dielectric 
dipoles the orientation by the applied field is opposed by thermal disorder. 

We first consider the order of magnitude of the permanent dipole/) arising 
from an orbiting electron. For a circular orbit of radius r and electron 
velocity v we have [see equation (14.10)] 

P = ^- (14.24) 

But the angular momentum mvr is quantized in units of — where h is Planck's 

constant. Putting 

I h\ 


where n is an integer, and substituting in equation (14.24) we have 

P = n (£^ . (14.26) 

The quantity in the bracket is known as the Bohr magneton. As a first 
approximation we expect to find certain atoms in which the angular momenta 
do not cancel out, so that one orbit (or one spin) is left unpaired. The 
magnetic moment of such an atom will be one Bohr magneton the magnitude 
of which is 



Since orbits, and spins, tend to occur in pairs of opposite momenta (or 
spins) we should not expect to find atoms with magnetic moments much 
greater than this. This has an important consequence which greatly eases 
further analysis. As we shall see below, at low temperatures and moderate 
fields H, it is possible to orient all the dipoles and produce magnetic satura- 
tion. Nevertheless, even at this stage, the magnetic moment Mper unit volume 

261 Magnetic properties of matter 

= "°{t)' 

is small compared with H. Consequently the difference between B and H, or 
between H and the true field seen by the individual atom, is negligible and 
can be ignored. In what follows, therefore, we consider simply the effect of the 
applied field H. 

Following the analysis for para-electric behaviour given in the previous 
chapter, we expect to find for small and moderate fields a constant sus- 
ceptibility given by 



For a solid occupying 10 c.c. per gram-atom, n = 6 x 10". Assuming p is one 
magneton we find that, at 300°K., x has a value 

X a 10-*. 

This is 50 to 100 times greater than the diamagnetic susceptibility. Conse- 
quently the diamagnetic susceptibility, which all atoms possess, is swamped 
by the paramagnetic susceptibility if the atom contains an unpaired orbit or 

For strong fields or low temperatures we obtain the corresponding Langevin 

5 [cotb^-^1 
H [ kT pH\ 


This is drawn in figure 101. Although this curve is identical with that obtained 
for para-electrics there are two basic differences between paramagnetic and 


*<.* 1-0 







f\ slope —i 





Figure 101. The Langevin function for paramagnetism, treating the orientation 
as a continuous (classical) phenomenon. 

262 Gases, liquids and solids 

para-electric materials. The first concerns saturation. Low temperatures 
clearly favour saturation in both cases. However most dielectrics possessing 
electrostatic dipoles become solids at low temperature and orientation 
becomes difficult or impossible; orientation is thus frozen-out. With para- 
magnetics we are only concerned with the orientation of orbits (or spins) 
and this can occur at the very lowest temperatures. Thus one can greatly 
enhance paramagnetism by operating at very low temperatures; one cannot 
in general do this with para-electrics. 

As an example of saturation with paramagnetics we note from figure 101 
that this occurs when pB/kT > 1, i.e., when 

H>—. (1430) 


Up = 10 _2 ° e.m.u. and T = 1°K. we obtain saturation when 

H> 1 -4x10* gauss. 

Such a field can be easily obtained and, with it, saturation can be achieved at 
temperatures of about 1°K*. 

The second difference concerns the nature of the polarizing field. Consider 
a typical solid occupying 10 cm. 3 per gram-atom; then n ~ 6 x 10 22 .Suppose 
all the dipoles are aligned by operating at 1°K. with a field H = 14,000 gauss. 
The magnetic moment per unit volume M - tip ~ 6 x 10 2 . The quantity 
AnM still remains appreciably less than H, so B is not greatly different from 
H. If one constructs a spherical cavity as in the dielectric case the quantity 
4nM(3 is less than 20 per cent of//. Thus the whole problem of deciding, 'what 
is the real field that the individual atom sees?' is relatively unimportant. We 
may simply assume that it is nearly equal to H without loss of accuracy. Of 
course if we carried out the experiment at 0-1 °K. so that saturation occurred 
for a field of only 1400 gauss, the problem would become more acute and we 
should have to carry out an elaborate analysis. In practice, the issue becomes 
important only in ferromagnetism. In that case, as we shall see below, the 
interaction due to the magnetic polarization of the material is so strong that it 
completely dominates the behaviour. 

The difference between paramagnetic and para-electric behaviour arises 
from the size of the unit dipole. In electrostatics it has a value of about 
4-8 x 10 _18 e.s.u.; in paramagnetism the Bohr magneton has a value of 
about 10 -20 e.m.u. The electrostatic unit dipole is thus about 500 times 
greater than the corresponding magnetic dipole. For this reason the interaction 
of the electrostatic dipole with the applied field is far more important. 

*This treatment is valid for gases and for paramagnetic salts where the individual dipoles play their 
full part. With paramagnetic metals the major contribution is from the spin of the free electrons and 
only those at the top of the Fermi level are able to take part, i.e. only a fraction of order kTIE*. 
seep. 191. ' 

Magnetic properties of matter 

14-4 Ferromagrjetism 

14-4-1 Theory of ferromagnetism 

Certain materials such as iron, nickel and cobalt show very high mag- 
netization M (typically M ~ 1000) for fairly modest fields even at room 
temperature. They reach magnetic saturation very easily. With an 'ordinary' 
specimen of iron this occurs for fields of the order of a few hundred gauss 
(see figure 102); with good single crystals in favourable orientations for a 
field of order only 1 gauss, so under these conditions x is about 1000. 






H gauss 

Figure 102. Magnetic moment M produced in a typical magnetic material for 
fields of order 1 000 gauss. 

A typical value of M at saturation is about 1000. If a gram-atom occupies 
10 ex. the number of atoms per c.c. is 6 x 10". At saturation, if these give a 
moment of 1000 e.m.u., each atom must contribute a magnetic moment of 

— — — ~ l-6xl0 -20 , i.e., about 1 Bohr magneton. This suggests that a 

saturated ferromagnetic consists of completely aligned dipoles. In this sense 
they are like paramagnetics, but with paramagnetics this result can only be 
achieved by using enormous fields or by operating at extremely low tempera- 
tures. With ferromagnetics alignment occurs with relatively weak fields. 

It was this conclusion that led A. Weiss to suggest that these materials are 
spontaneously magnetized over fairly large domains and that these domains 
are easily aligned when an external field is applied. There is a co-operative 
interaction between the atoms so that an additional field IM acts on each 

atom. This resembles the Lorentz field in dielectrics | — P \ , but the quan- 
tity A is immensely larger, of the order 10 4 . The resultant field producing 

264 Gases, liquids and solids 

magnetization may then be written 


(or B + X'M where X' = X-4n). 
The Langevin equation for the alignment of the dipoles becomes 

M = np cothx — 

where now 

x = -£- [H+XM]- 




The behaviour of the Langevin equation as a function of H is best seen by 
rewriting equation (14.32) as 

— = coth* — , 
np x 

and equation (14.33) as 

M_ xkT H 
np p 2 Xn Xnp 



Consider first the behaviour when H = 0. This corresponds to the spon- 
taneous magnetization of ferromagnetic materials in the absence of an 

applied field. We plot equation (14.34) as — against x and obtain the 


typical curve of figure 103(a). The slope near the origin is \. In figure 103(6) 

we have plotted equation (14.35) as — against x, for H = 0. We obtain 






equation (34) 



^equation (35) 








S / 

/ / 

/ / 


i / / 




Figure 103. Graphical method of explaining ferromagnetism. (a) Graph of 
equation (14.34). (b) Graph of equation (14.35). (c) Superposition of both 
graphs. For curve (i) there is no intersection, i.e. ferromagnetism does not occur. 
For curve (ii) intersection implies that ferromagnetism does occur. 

266 Magnetic properties of matter 

a straight line of slope — — . Clearly if this slope is greater than + no 

solution is possible and there is no magnetization [figure 103(c), curve (i)]. 

If it is less than \ magnetization and saturation are easily achieved [figure 
103(c), curve (ii)]. 

It is at once evident that there is a critical temperature above which no 
ferromagnetization will occur. This is known as the Curie temperature and 
is given by 

kT c 
np 2 X 


If the saturation value of M is written as M (M = np) the drop-off of M 
with increasing temperature is shown in figure 104 as a plot of MjM against 









\ diamagnetism 


\ and 

\ paramagnetism 

i i i i 


0.4 0.6 


Figure 1 04. Drop in magnetic moment of a ferromagnetic material as temperature 
is raised. Above the Curie temperature 7c the material shows only diamagnetism 
and paramagnetism. 

T. Above T c only paramagnetism and diamagnetism remain. In what follows 
we shall ignore the diamagnetism and calculate in a very simple way the 
paramagnetism above T c . If the fields are not too strong the Langevin 
equation (14.34) may be written 




Similarly, using equation (14.36), equation (14.35) may be written 

M m xT_H_ 
np 3T e hip ' 



266 Gases, liquids and solids 

Substituting for x from equation (14.37) we obtain 

np \T C J Xnp 
Consequently the paramagnetic susceptibility becomes 

= M = 1 / T c \ 
X H X\T-T C )' 

From equation (14.36) this becomes 

X = 





3k(T-T € ) 

We see at once that X has disappeared explicitly although, of course, it 
is implicit in the value of T c . Equation (14.41) is well obeyed and gives good 
quantitative agreement for most ferromagnetic materials on the assumption 
that each atom contributes of the order of 1 or 2 Bohr magnetons. 

Finally we may note that in the ferromagnetic range an applied field has 
little effect compared with the internal field. Even for a field H =: 10* gauss 

the interval — shown in figure 105 is small and has very little effect on the 

resultant magnetization. 



/ - 



" kT 1 


Figure 105. An applied field H has little effect on the magnetization of a ferro- 
magnetic material. 

The origin of the internal field: ferromagnetism and anti-ferromagnetism 

It is not possible to explain the origin of the large internal field in simple 
terms. Indeed it introduces very difficult quantum mechanical problems. 

267 Magnetic properties of matter 

We may, however, describe it in the following terms. The large value of A 
arises from the 'exchange' interaction between the atomic dipoles. This 
interaction depends on the ratio of the orbital radius r to the distance a 
between neighbouring atoms. Over a critical range (air about 3) the mag- 
netic exchange interaction is large and positive. This implies that for atoms 
with large dipoles and of the correct air ratio strong positive interaction 
will occur, A will be large and the material will be ferromagnetic (Fe, Co, 
Ni). This is shown schematically in figure 106. It is also seen that for small 

Figure 106. Exchange interaction in ferromagnetic materials as a function of -. 

values of a/r the interaction can be negative. This favours an anti-parallel 
orientation of neighbouring spins and materials showing this property 
(MnF 2 , MnO, CoO) are known as anti-ferromagnetic substances: at low 
temperatures the interaction is effective and an applied field produces little 
magnetization so that the susceptibility is small. As the temperature is raised 
the anti-ferromagnetic interaction becomes less effective and the susceptibility 
of the material increases. Above a critical temperature known as the Neel 
temperature the interaction is completely swamped by thermal motion. The 
material ceases to be anti-ferromagnetic and becomes paramagnetic in the, 
more or less, classical sense. The detailed behaviour is very complicated and 
we shall not discuss it further. 

14 -5 Quantum treatment of magnetic properties 

The classical treatment of diamagnetism is satisfactory from two points of 
view. First the concept of induced currents producing fields which oppose the 
applied field seems reasonable and tractable. Secondly it gives the right 
answer for the diamagnetic susceptibility. From all other points of view it is 

268 Gases, liquids and solids 

Detailed considerations, first described by N. Bohr in 1911, show that if 
classical mechanics are applied rigorously to an orbiting electron the para- 
magnetic and diamagnetic terms cancel one another exactly so that there is no 
net induced magnetization. The reason is briefly as follows. Assuming that an 
electron orbiting around a nucleus is a stable classical system the radius of a 
given orbit can have a continuous range of values from to co rather than the 
one particular size assumed in the Langevin theory. Further, whereas the 
Langevin theory applies the Boltzmann energy distribution to the external 
degrees of freedom of the orbiting electron, that is to the orientation of the 
orbits in paramagnetism, it does not do so to the internal degrees of freedom, 
that is to the orbital energies of the electron in its treatment of diamagnetism. 
This is logically inconsistent. If the Boltzmann distribution is applied to both 
orientation energies and orbital energies the paramagnetic and diamagnetic 
parts exactly compensate one another. The Langevin treatment gives the 
correct answer because it assumes a finite radius and an orbital energy 
independent of temperature; in a sense it incorporates hidden quantum 
conditions although in fact Langevin's theory (1905) preceded Bohr's 
quantized atom (1913) by eight years. 

Classical electromagnetic concepts cannot really be applied to a classical 
atom since a non-quantized orbiting electron constitutes an unstable system 
(see below). The only situation for which such concepts can be applied 
meaningfully is to a free 'electron-gas', for example the electron plasma in a 
metal. Here again, the effect of a magnetic field is to produce diamagnetic 
and paramagnetic effects which exactly cancel one another. The reason is 
simply that in classical electromagnetism a field can deflect an electron but 
it cannot do work on it,* so that although the electrons are deflected into 
helical paths their energy distribution remains unaltered. This implies that 
the magnetic field produces no energy change in the system, that is the system 
behaves as though it has no net magnetic property. A discussion of this issue 
is given in a very readable and interesting form in J. H. van Vleck (1932), 
The Theory of Electric and Magnetic Susceptibilities, Oxford, Clarendon Press. 

As mentioned above the classical atom is unstable; the orbiting electrons 
would spiral into the nucleus radiating all their energy in the process. The 
effective lifetime of such an atom would be infinitesimally small. Atoms exist 
as stable entities precisely because energy states (or angular moments) are 
quantized. To produce a perturbation of the energies by the classical methods 
is quite invalid. However, quantum mechanics shows that there is a genuine 
diamagnetic effect which can be calculated rigorously. The analysis is far too 
advanced to be given here. It gives the same answer as the classical treat- 
ment which we have given in this chapter in section 14-2. For this reason, as 
in most other elementary texts, we have persisted in giving the simple treat- 
ment: the answer is right though the physics is not entirely satisfactory. 

* Of course a magnetic field exerts a force on a conductor through which a current is passing. If the 
conductor moves, work is done, but the energy does not come from the magnetic field. The energy is 
provided by the electric current flowing through the conductor in overcoming the back e.m.f. generated 
in the conductor itself as a result of its movement in the magnetic field. 

269 Magnetic properties of matter 

We have taken over from our chapter on dielectric properties the classical 
treatment of paramagnetism. With para-electrics quantum effects play a 
trivial part and the classical treatment is perfectly satisfactory. This is not so 
with paramagnetics. The reason is that the individual dipole can take up only 
a discrete number of orientations. If for example an atomic system has a 
resultant angular momentum represented by a vector of length f, the only 
orientations which the system can acquire in a magnetic field correspond to 
components of f , \, —\ and — § in the direction of the field. In classical 
physics the orientations would cover the whole continuous range such that 
the components have values from +f to — f . The effect on paramagnetism 
is not very marked but it is worth considering the simplest possible case when 
the dipole of the individual atom is due to a single electron spin. The only 
orientations allowed are those which are exactly parallel and anti-parallel to 
the applied field. If there are n atoms per c.c, and of these n+ are those with 
orientations along the field and «_ those with orientations opposing the 
field, we have by the Boltzmann principle 

« + = A exp (^\ (14.42) 

n_=^exp/-^. (14.43) 

Since n+ + //_ = n we find that 

—K-?) + " p (-*-?)P 

The net moment per unit volume is then 

M = («+ - n.)p (14.45) 


= np± — 

= np tanh \*—\ . (14.46) 

In form this curve is very similar to the Langevin function. However there is a 
quantitative difference in the region of small fields. Since for small values 
of x, tanh x ^ x we have from equation (14.46) for small values of H, 

l P H\ np 2 H 
The susceptibility is then 

_ M _ nj? 
X ~ H~ kT' 

270 Gases, liquids and solids 

This is just three times the value deduced from the Langevin function. Similar 
corrections may be applied to the discussion of ferromagnetism and the Curie 
temperature given above. However the correction factor of 3 will only apply 
rigorously for a single-electron-spin system. With most materials the situation 
is usually more complicated and in fact the correction factor becomes less 
than 3. 

14-6 Ferromagnetic domains 

We consider here briefly the magnetization of a single crystal of a ferro- 
magnetic material such as iron. With iron the 6 cube edge directions [100] 
are the directions of easy magnetization. If a magnetic field H is applied 
along the edge direction the domains are aligned easily and saturation is 
achieved for a very small field. If the field is applied along, say, the face 
diagonal direction [1 10] the easy orientation directions are at 45° to the field., 
Consequently for small fields the net magnetization in the direction of H is 
l/v/2 of what it would be if the field were applied along the [ICO] direction. 
Experiments show that this is so (see figure 107 ). If the field is increased 
further magnetization can occur only by twisting the domain directions out 
of their easy orientations. This is a far more difficult process and relatively 
strong fields must be used to achieve it. 

Figure 107. Magnetization of a single crystal of iron along a cube direction [100] 
and along a face diagonal direction [1 1 0]. 

14 -7 Magnetic hysteresis 

When a magnetic field is applied to a magnetic material, the magnetization 
generally increases as the field is increased. If the field is reduced, the 
magnetization diminishes but does not follow the original curve - it 'lags 
behind'. This effect is known as magnetic hysteresis. 

For the following simple description of magnetic hysteresis I am indebted to 
Dr Shoenberg. Consider the behaviour of a material which has a single easy 

Magnetic properties of matter 





Figure 108. Magnetization of material showing a single direction of easy mag- 
netization, (a) Initial unmagnetized state, (b) Effect of applying a weak field to the 
right is to orient all the domains in that direction. This gives magnetic saturation ; 
on removing the field the magnetization remains unchanged so that the remanence 
is almost equal to the saturation magnetization, (c) Magnetization curve showing 
that the remanence is very large and that demagnetization can be achieved only 
by reversing the sign of the field. 

direction of magnetization. In the unmagnetized state as many domains are 
magnetized to the left as to the right [see figure 108(a)]. If a field is applied 
to the right the domains all become magnetized to the right and the material 
is saturated [figure 108(6)]. The work done is used in moving the domain 
walls surrounding the right-pointing domains, until they swallow up the left- 
pointing domains. If the field is now removed no further change occurs and 
the material shows 100% remanence [see figure 108(c)]. The magnetization 
can be reduced only by applying a strong field to the left. 

In general metal specimens contain internal stresses. If the stresses are 
small they merely decide which of the easy directions of magnetization are 
easiest of all. If the stresses are large they may completely dominate the 
behaviour; in that case the easy domain directions are determined by the 







\- i / 

T J // 

T /\- 

\- I / 
/ T -*S 

I !// 

Figure 1 09. Magnetization of a material showing a limited number of directions 
of easy magnetization (vertical, horizontal and at 45°). (a) Initial unmagnetized 
state, (b) a horizontal field H aligns all the domains, (c) on removing the field tha 
domains revert to their original directions but the sense is that of the field H. Undsr 
that* condition* th9 remanence it about half the saturation magnetization. 

272 Gases, liquids and solids 


stresses rather than by crystal orientation. Consider for simplicity a specimen 
in which, as a result of local stresses, the domain directions are either hori- 
zontal, vertical or at 45° to these directions [figure 109(a)]. If a strong 
horizontal field is applied all the domains line up with the field and the material 
achieves saturation [figure 109(6)]. If the field is now removed the domains 
return to the original orientations they had in figure 109(a), but with the 
sense imposed on them by the applied field. This is shown in figure 109(c). 
Under these conditions the material has a remanence of about half the 
saturation magnetization. 

Summary: Comparison of dielectric and magnetic properties 

We now summarize in note form the main points of resemblance and differ- 
ence in dielectric and magnetic properties. 
Dielectric Magnetic 

Electronic polarization. Distortion 
of electron cloud. Critical frequency 
about 10 15 . Induced dipole: 

p = r 3 E 

Always in direction of E. 

Diamagnetism. Deceleration or ac- 
celeration or precession of those 
electrons possessing orbital momen- 
tum. Susceptibility per electron: 

ie 2 r 2 
6 mc 1 

Always opposes field H. 

Unaffected by temperature. 

Atomic polarization; stretching or 
bending of bonds if molecule has 
permanent dipole. Critical fre- 
quency about 10 12 . 


Para-electric: orientation of mole- 
cular dipole p by applied field E, 
opposed by thermal motion. Typi- 
cal value of p = 5xl0 -18 e.s.u. 
For small E, susceptibility per mole- 
cule is: 

Paramagnetic : orientation of atomic 
dipole by applied field H, opposed 
by thermal motion. Typical value 
of p = 10" 20 e.m.u. For small H, 
susceptibility per atom is: 

J— (classical) 


—~ (quantum mechanical) 

273 Magnetic properties of matter 

Para-electric polarization is of order 10 3 larger 

than paramagnetic. Hence need to consider real 

field rather than applied field. 

Very markedly dependent on temperature. 

Full behaviour described by Langevin function. 

Para-electric behaviour cannot be Paramagnetic effects can be greatly 

easily enhanced by low tempera- enhanced by low temperatures 

hires because dielectric solidifies since only orbits or spins have to be 

and molecules cannot be oriented. oriented. 

Ferroelectric Ferromagnetic 

Co-operative phenomena in which internal field is augmented by some type 
of specific interaction. Large oriented domains of permanent dipoles. Above 
some critical temperature (Curie temperature) the interaction is ineffective, 
and the materials cease to be ferroelectric or ferromagnetic. 

274 Gases, liquids and solids 


275 Appendixes 

Appendix A 

c.g.s. and MKS units 

The ratio given in the right-hand column is the factor by which the quantity 
in the MKS system must be multiplied to convert into the c.g.s. system. 


Unit in MKS 

Unit in c.g.s. 

_ ,. MKS 







metre m. 

centimetre cm. 

10 2 


kilogram kg. 

gram g. 

10 3 


second sec. 

second sec. 



kg. m.~ 3 

g. cm. -3 

10" 3 


m. sec. -1 

cm. sec."' 

10 2 


m. sec. -2 

cm. sec." 2 

10 2 


kg. m. sec. -1 

g. cm. sec." 1 

10 s 


newton N. 


10 5 


newton metre 

dyne cm. 

10 7 

Work, energy 

N. m. = Joule J. 


10 7 


watt: J. sec. -1 

erg sec. - ' 

10 7 

Surface tension 

N. m." 1 J. m." 2 

dyne cm."'; 
erg cm." 2 

10 3 


kg. m. -1 sec."' 

g. cm."' sec."' 


276 Gases, liquids and solids 

Electrical and Magnetic RMKS 


Unit in MRS 

Unit in c.g.s. 

_ . MKS 




Charge q 

coulomb C. 



Current i 

ampere A. 



Potential V 






Electric field E 

voltm. -1 , N.C. -1 


ixlO" 4 


polarization P 

Cm." 2 


3xl0 5 

Inductance L 



10 9 

Resistance R 




Capacitance C 





induction B 

weber m.~* 



Magnetic field H 

ampere m. -1 


4tix10 3 

Intensity of mag- 


netization M 

weber m.~ 2 




io 10 

moment m 

weber m. 





susceptibility x 

MKS m." 3 

e.m.u. cm. -3 

Wherever a factor 3 appears in the electrical and magnetic conversion factor, 
it implies the approximation for the velocity of light c — 3 x 10 10 cm. sec. The 
basic definition of fio in RMKS is 4nx IO -7 henry m. -1 and the permittivity 
is e = (Moc 2 )' 1 (see B. I. Bleaney and B. Bleaney (1911), Electricity and 
Magnetism, Clarendon Press). 

277 Appendix A 

Appendix B 

Values of some physical constants 



Value c.g.s. 

Value MRS 

Charge on electron 


4-803 x 10- 10 e.s.u. 

l-602xl0- 20 e.m.u. 

l-602xlO- 19 C. 

Mass of electron 


9-108 x 10"" g. 

9-108xl0" 31 kg. 

Electron charge: 


5-273 x 10" 17 



e.s.u. g. _1 

l-759xl0 7 

l-759xl0 11 1 . 

e.m.u. g.- 1 

Boltzmann constant 


l-380xl0- 16 

1-380 xlO" 23 

ergdeg. -1 

joule deg. -1 

Avogadro's number 


6025x10" mole" 1 

6025xl0 23 mole" 1 

Gas constant 


8-317X10 7 

8-317 joule mole -1 

erg mole -1 deg." 

deg. -1 

1-967 cal. mole -1 

deg. -1 



4186 xlO 7 


equivalent of heat 

erg cal. -1 

4-186 joule cal. -1 


Vol. 1 mole gas at 

22-420xl0 3 cm. 3 

22-420 x 10" 3 m. 3 


Faraday constant 


2-894xl0- 13 
e.s.u. mole -1 

96,520 C. mole" 1 

Planck's constant 


6-625X10" 27 

6-625 xl0- s * 

erg sec. 

joule sec. 

( h \ 


1054 xlO" 27 

1054xl0- 34 


erg sec. 

joule sec. 

Velocity of light 


2-998 xlO 10 
cm. sec. - * 

299800 km. sec." 1 

278 Gases, liquids and solids 


Value c.g.s. 

Value MKS 

Ratio proton: 
electron mass 





Bohr magneton 



9-273 xlO- 21 
erg gauss -1 

1-I65xl0- 29 
weber m." 1 

Energy equiv. of 1 eV 

1-602x10- "erg 
2305 kcal. mole" 1 

l-602xl0 -19 joule 

Gravitational constant 


6-67 x 10" 8 
dyne cm. 2 g. -2 

6-67 xlO" 23 
N.m. 2 kg." 2 

Permeability of free 



47ixl0 -7 henry 
m." 1 (RMKS) 
10" 7 MKS 

Permittivity of free 

e = 
(AoC 1 )- 1 


8-854 10" 12 farad 

m." 1 (RMB^S) 

1113 lO" 10 MKS 

The values in the above table for N , R, /and ^are 'physical' constants based 
on the number of atoms in 16 g. of 16 0. The equivalent constants used by 
chemists are sometimes based on a different gram-molecular convention and 
the values quoted may differ by up to one part in a thousand. 

279 Appendix B 


Activation energy, for viscous flow 

Adhesion, and atomic forces 16 
Adiabatic expansion 

of imperfect gases 1 1 6-2 1 
gas-kinetic, theory of 87 

of perfect gases 
irreversible 34 
reversible 36-8 
Adiabatic lapse rate 40 
Ageing, role of, in upper yield 

phenomenon 163 
Alpha particles, role of, in determining 

No 5 
Andrews, experiments of, on C0 2 98 
Angle of contact 215, 219 
Angular distribution 

in kinetic gas theory 66 

in orientation of dipoles 242, 270 
Anomalous dispersion, theory of 246 
Anti-ferromagnetism 268 
Atmosphere, law of 40, 74-5 

and Boltzmann distribution 75-7 

density of 12 

forces between 
attractive 13-16 
repulsive 16-18 

periodic table of 10 

potential energy of 18-19 

size of 10,11 

structure of 12 
Attraction of walls, for gas molecules 

Attractive forces 

covalent 13 

gravitational 21 

ionic 13 

van der Waali 16-19 

Avogadro's law (hypothesis) 4, 34 

kinetic theory of 36 
Avogadro's number 

definition of 4 

determination of 5-7 

Beams, bending of 136-8 
Bernal, model of liquid state by 

Bernouilli's equation, for ideal-liquid 

flow 223^4 
Berthelot's equation of state for 
imperfect gases 105 
critical points of 1 14 
Bingham fluid, definition of 234 
atom 269 
magneton 261 
Boiling point, relation of, to critical 

temperature 209 
Boltzmann distribution 76 
application of 
to collisions of gases with walls 

to orientation of dipoles in a field 

242, 270 
to reaction rates 76 
to vapour pressure 208-10 
to velocities in a gas 78 
to vibrations in a gas 89 
Boltzmann's constant 16 
role of 
in Brownian motion 93 
in critical temperature 100 
in equipartition of energy 92 
in specific heats 85-92 
Bond energy 146-7 
Bora's theory of liquid stat* 202 

281 Index 

Boundary conditions 
in flow of gases 53-5 
in thermal conductivity 57 
Boundary lubrication 228 
Boyle's law 
kinetic theory of 45 
statement of 33 
Boyle temperature 
definition of 98 

derivation of from van der Waals 
equation 107 
Brinell hardness test 157 
Brittle failure 154 
atomic mechanism of 166 
criterion of 165 
role of cracks in 167 
Brittle solids, mechanism of rendering 

ductile 169 
Brownian motion 
comments on 
by Lucretius 95 
by Perrin 96 
theory of 93-6 
Bulk modulus 
definition of 135 
theory of 
ionic solids 140-43 
metals 147 
van der Waals solids 144-6 

Cantilever, bending of 136-8 
Capillary rise 

energies of 221-2 

theory of 217 
Carnot heat engine, efficiency of 

Charles' law 

kinetic theory of 46 

statement of 33 

concept of entropy 3 1 

equation of state 105 
critical points of 114 
Clausius-Clapeyron equation 210-11 
Clausius-Mosotti equation 251 
Cloud chamber 220 
Cohesion, in solids 165 
Collision frequency, in gas kinetic 

theory 70 

Collisions with wall, deductions of, 

using Boltzmann distribution 82 
Colloidal systems, role of van der 

Waals forces in 16 

see Electrical Conductivity and 
Thermal Conductivity 
Contact angle 215, 219 

on adiabatic expansion 38 

on Joule-Kelvin expansion 118 
Corresponding states, law of 115 
Coulombic attraction, between ions 

13, 140 

dislocation pile-up, as cause of 169 

effectiveness of, as stress-raisers 

Griffith theory of 168 

role of, in brittle fracture 167 
Creep, in solids 198 
Critical coefficient, as deduced from 

Bert helot's equation 114 

Dieterici's equation 114 

van der Waals equation 114 
Critical temperature 

definition of 99 

derivation of 108-13 

intermolecular significance of 99 

values of 100 
Critical velocity, in liquid flow 225 
Crystalline solids 

structure and bonding of 122-3 

see also Surface energy, Bond 

energy and Bulk modulus 

Curie temperature in ferromagnetism, 

theory of 266-7 
Curvature, and surface tension forces 



atomic theory 4 

law of multiple proportions 3 

law of partial pressures 34 
kinetic theory of 46 
Debye temperature, values of 178 
Debye theory of specific heats of solids 


282 Gases, liquids and solids 

Defects, role of, in strength properties 
of solids 

cracks 166 

dislocations 161 

general 170 
Degrees of freedom in specific heat 

calculations 85-93 
Democritus, views of, on hardness and 

density 3 

definition of 256 

Langevin theory 258-61 

quantum theory 269 
Dielectric constant 

definition of 235 

relation of, with polarization 235-6 

relation of, with optical properties 

theory of 237-53 
Dielectric constant of water and ice, 

variation of, with frequency 252 
Dielectric properties of gases 

comparison of, with critical data 239 

effect of alternating field on 245 

relation of, with optical properties 
Dielectric properties of liquids and 
solids 248-9 

strong interactions in 249-52 

equation of state for imperfect gases 
critical points of 114 

of gases in gases 57-60 

of impurities in solids 163 
Diffusivity, thermal 186 

electrostatic, comparison of with 
magnetic 263 

orientation of 
in electric field 242-5 
in magnetic field 262, 270 
Directional distribution, in gas theory 


edge, role of in shear 161-2 

experimental evidence for 162 

screw, role of in crystal growth 164 

Dislocation pile-up, as cause of cracks 

Dispersion, optical, relation of, with 

dielectric properties 246-8 
Displacement vector, definition of 237 
Distributions in gas theory 

directions 65 

velocities 77-85 
Drude's electrical conduction theory 

Ductility 154-5 
Dulong and Petit's law 30 
Dupre equation 215 

experimental study of 216 

Einstein temperature 173 

relation of, with Debye temperature 
Einstein theory 

of Brownian motion 93-5 

of specific heat of solids 172-3 
Elastic moduli 

definition of 134-6 

relations between 135 
Elastic properties of crystalline solids 

elementary theory 131-2 

general behaviour 154 

lattice theory 
ionic 140-43 
metals 147-8 
van der Waals 144-6 
Elastic properties of rubber 148-53 
Electrical conductivity of metals 

Drude model of 188-90 
Electrical noise 93 
Electrical properties of solids 1 88-92 
Electrolysis, use of, in determining 

No 6 
Electron gas 43, 188-190 
Electron spin, role of 

in atomic structure 8 

in magnetism 270 
Electrostatic dipole. definition of 241 
Energy, calculation of 

bond (lattice) 146-7 

rotational 88, 91 

vibrational 88, 89 

283 Index 

Energy, equipattition principle of 
on macroscopic scale 92-6 
on molecular scale 85-92 

Energy, free 
Gibbs 32 
Helmholtz 32 

Energy, internal 25-8 

Energy, surface 212-14 

Energy, potential, between atoms and 
molecules 17-22 

Energy, role of 
in ether theory of gases 43 
in kinetic theory of gases 42, 87 

Enthalpy 119-21 
properties of 28 
role of. in Joule-Kelvin expansion 

constancy of, in adiabatic expansion 

definition of 31 
relation of, with probability 32 
role of, in rubber elasticity 152 

Equations of state, of imperfect gases 

Esterman, Simpson and Stern, 
experimental results of, for deter- 
mination of velocity distributions 

Ether theory of gases 43 

of droplets 220 
kinetic theory of 208 
role of, in diffusion 59 

Exchange interaction 
in ferromagnetism 267-8 
in molecular binding 13 

hole model of liquid 199 
theory of liquid viscosity 230-33 

Faraday 6 

Fermi distribution 190 

Fermi electron, specific heat of 

Ferroelectricity 252 

domains in 271 
hysteresis in 271-3 
quantum theory of 270 


Ferromagnetism - contd 

saturation in 264 

Weiss theory of 264-7 

monolayers of, on water 5 

soap, tension of 217 
First law, of thermodynamics 26 
Flexure, of beams 136-8 

in ideal liquids 223 

in real liquids, Newtonian 225 

in real liquids, non-Newtonian 234 

Poiseuille 226 

turbulent 225 

theory of (gases) 50-5, 68-70, 228 
theory of (liquids) 230-33 
Fluids, internal energy of 27 
Flux, magnetic 254, 258 
Forced vibrations, role of, in optical 

dispersion 246-8 
Forces, attractive 

ionic 13, 140-43 

metallic 147 

van der Waals 13-16, 144-6 
Forces, consequences of in solids 

Forces, repulsive 16 
Frank, mixture model of water by 206 
Free electron, specific heat of 188 
Free energy 

Gibbs, definition of 32 

definition of 32 
use of in rubber elasticity 152 

of atomic vibrations 140, 172 

of lattice waves 173-9 

relevance of, in Joule equivalence 25 

role of plastic work in 161 

general 199 

latent heat of 206-8 

temperature of 207 

Gas expansions 
entropy charges associated with 

284 Gases, liquids and solids 

Gas expansions - contd 

adiabatic 36-8 
definition of 27 
isothermal 34 
Gas velocities 
equations for 81 
typical values of 43 
bulk properties of 33-41 
. ether theory of 43 
kinetic theory of, advanced 65-85 
kinetic theory of, simple 41-64 
thermal conductivity of 55 
thermal equilibrium in 45 
viscous flow of 
advanced 68-70 
simple 50-5, 228 
Gibbs free energy, definition of 32 
structure and bonding of 122-3 
viscosity of 123 
Gravitational forces, between atoms 22 
Green's theory of liquid state 202 
Griffith theory of crack propagation 

Gruneisen constant, definition of 182 
Gruneisen theory of specific heats 


of brittle solids 170 

of ductile solids 157 

concept of 24 

flow of, along solid bar 1 84 
Heat of sublimation in solids 

table of values of 126 

theory of 125 
Heat transfer coefficient 57 
Helmholtz freeenergy, definition of 32 
Hooke's law for elastic deformation 

Hund's law, in atomic theory 8 
Hydrostatic pressure, role of 

in brittle failure 165-7 

in elastic compression 134 

in hardness measurements 157 

in plastic deformation 155 

Hydrodynamk lubrication of shaft 
in bearing 226-7 


dielectric properties of 252 

specific heat of 178 

structure of 204 

transformation of, on melting 
Ideal gas laws 33-41 
Imperfect gases 

equations of state for 111-15 

virial equation for 97 
Induction, magnetic 254 
Interfacial energy, for liquid on solid 

215, 222 
Intermolecular forces 19-21 

role of, in imperfect gas 104-8 
Internal energy 

of gases 86 

general thermodynamic properties of 

of rubber 152 

of solids 141-8,171-9 
Internal field in ferromagnetism, 

origin of 267-8 
Inversion point, in atmospheric 

temperature 40 
Inversion temperature, in Joule-Kelvin 

theory of 118—19 

values of 121 
Tons, forces between 13, 140 

size of 
in a linear chain 143 
in a simple molecule 143 
in a solid crystal 12, 143 
Isoenthalpic behaviour, in Joule- 
Kelvin expansion 119-20 
Isothermal behaviour 

of ideal gases 34, 37 

of imperfect gases 97-9 

Jeans, treatment of, for excluded 

volume in gases 103 
Joule, experiments of, on mechanical 

equivalent of heat 24-6 
Joule expansion into vacuum, of ideal 

gas 34-5 

285 Index 

Joule-Kelvin expansion of imperfect 
gas 116-18 
inversion temperature in 118-19 
role of enthalpy, in 119-21 

Kay's study of Dupre equation 216 

see Joule-Kelvin 

Kinetic theory of imperfect gases 

Kinetic theory of perfect gases 
advanced 65-96 
elementary 41-64 

Ko's determination of velocity dis- 
tribution in gases 84 

Langevin function 244, 262 
Langevin theory 
of diamagnetism 258-61 

inadequacy of 269 
of paraelectricity 242-5 
of paramagnetism 261-3 
Lattice waves, role of, in Debye 

theory 174-9 
Liquid meniscus 
pressure difference across 216-17 
vapour pressure over 220 
Liquid state 
continuity of, with gaseous state 99 
miscibility of, with other liquids 100 
model of 

as modified gas 193-6 

as modified solid 199 

sui generis 199-204 
flow of 

ideal 223 

real 225-8 
rigidity of 233 
surface energy of 212 22 
structure of 

Alder 202-4 

Bernal 199-202 

Born and Green 202 

Eyring 199, 206 

general 196-8 

Moelwyn-Hughes 199 
tensile strength of 196 
vapour pressure of 208-12 
viscosity of 228-34 

Lorentz, treatment of, for polarization 

in condensed state 249 

boundary 228 

elastohydrodynamic 228, 232 

hydrodynamic 226 — 8 
Lucretius, quotation from, on Brownian 

motion 95 

Magnetic equations and definitions 

Magnetic induction 254 
Magnetic hysteresis, mechanism of 

Magnetic permeability, mechanism of 

see also Anti-ferromagnetism, 
Diamagnetism, Ferromagnetism 
and Paramagnetism 
Magnetization of single crystals 271 
Matter, particulate nature of 3 
Mean free path 

of conduction electrons 1 88 

of Fermi electrons 192 

of gas molecules 46-8, 70-72 

of gases at critical point 113-14 
Mean molecular separation in gases, 

comparison of, with molecular 

diameter 48, 113-14 
Mean square displacement, in 

Brownian motion 94 
Mean square velocities in gases 

advanced theory of 77-81 

simple theory of 41-2 

typical values of 43 
Melting, Lindemann theory of 207-8 
Membrane, semipermeable, in vapour 

pressure studies 211 

bifurcation of 129-30, 137 

surface energy of 129 
in presence of liquid and vapour 
Millikan's determination of electron 

charge 6 

molecular, in liquid state 198, 203 
solids 163 

288 Gases, liquids and solids 

Moelwyn-Hughes' model of liquid state 


collisions of gas on walls 49, 72-4 

dimensions 4. 60 

dynamics, in theory of liquids 

energy, thermal 85-92 

forces 19-22 

mean separation, in gases 48, 113 

softening, in gases 48-9 
Monatomic molecules 10 
Most probable velocity, in kinetic 

theory 81 
Multiple proportions, law of 3 

Neel temperature 268 

Neutral axis, in cantilever bending 

Neutron, in nuclear structure 8 
Nucleation, in condensation 220 
Nucleus, in atomic structure 8-10 

Oil drop experiment 6 

Optical dispersion, relation to dielectric 

properties 246 
Orbital magnetic moment 258, 261 
Order, short range, in liquids 206 

of dipoles in a field 
electric 242—5 
magnetic 262. 270 

of surface molecules 5 

in diatomic molecules 91 

quantized, energy of 89 

in solids 172-84 


effect of 
frequency 245-8 
temperature 248 

Langevin theory 242-5 

definition of 256 

effect of temperature 263 

Langevin theory 261-4 

quantum theory 268-9 

saturation in 263 

Pauli exclusion principle 

in atomic structure 8 

in repulsive forces 16 
Perfect gases 

advanced theory 33-60 

simple theory 65-96 
Periodic table 10-1 1 
Permeable membranes 21 1 
Permeability, magnetic 

definition of 254 

theory of 258-74 
Perrin, comments of, on Brownian 

motion 96 
PetrolTs bearing-equation 227 
Pockels' determination of molecular 

size 4 

definition of 187 

role of, in thermal conduction 187 
Planck's constant 88-9Z 172-9 
Poiseuille's equation, for viscous flow 

Poisson's ratio 

definition of 135 

value of, for rubber 148 
Polar molecules in electric fields 

bond deformation of 241 

orientation of 
in alternating field 245-6 
in static field 242-5 

electric, definition of 235 

general theory of 237-53 

of neutral atoms 238 

of polar molecules 242-5 
Potential energy curves 

between atoms 18 

between molecules 22 

relation to critical temperature 99 
Precession, of orbits in diamagnetism 

Pressure, effect of, on vapour pressure 

Pressure defect, in imperfect gases 

Preston's discussion of viscosity 123 
Principal radii of curvature, in surface 

tension calculations 217 
Protons, in nuclear structure 8 

287 Index 

Quantum theory 
of atomic structure 8-1 1 
of magnetic properties 268-71 

Radial distribution function 

definition of 197 

in liquids 197, 201 

in solids 197 
Radioactive disintegration, use of in 

determining N 6 
Random walk 149-52 
Range of molecular action 14-16, 

106. 165, 196,213 
Range of surface forces 222 
Rayleigh's determination of molecular 

size 5 
Rees's model of the liquid state 199 
Refractive index, theory of 

for gases 239-40 

for solids 249 
Relativity, effect of, on molecular mass 

Relaxation time, in liquid flow 233 
Remanence, magnetic, mechanism of 


see Electrical conductivity and 
Thermal conductivity 
Resonance effects in polarization of 

dielectrics 246-8 
Reynold's number, definition of 225 
Reynolds, hydrodynamic theory of 


modulus of, for solids 135 

of liquids 233 
Rock salt 

bond (lattice) energy of 146-7 

elastic modulus of 140-43 

structure of 7 
Rotational energy of molecules 88, 91 
Rotational freedom, in ice 178, 252 

elastic properties of 148-53 

structure and bonding of 122-3 
Rutherford's determination of 

Avogadro's number 6 

Sambursky, quotation from, on 
Lucretius 96 

Scott's model of the liquid state 

Screw dislocations, role of, in crystal 

growth 164 

of gases 74 

of particles in solution 75 
Self-diffusion in gases 57 
Semi-permeable membrane 21 1 
Shaft and bearing, lubrication of 

Shear stress 

calculation of 158-60 
comparison of with experiment 

role of dislocations in 161-2 

definition of 155-7 

effect of, on yielding 157 
Short range order, in liquid state 

197, 200 

of atoms 11 

of ions 12 

of molecules in imperfect gases 
Simpson 85 

Slip at boundary, gases 53 — 5 
Solid-liquid transitions 124-5 
Solio-liquid-vapour, equilibrium 

curves of 125 

bond energy of 146 

bonding in 122 

bulk modulus of 140-48 

elasticity of 131-8 

electrical conductivity of 188-92 

rotations in 178,252 

strength properties of 154-70 

sublimation of 125 

surface energy of 127-31 

thermal conductivity of 184-8 

thermal energy of 171-9 

thermal expansion of 132, 179-84 

vibrations in 138, 172, 174 

wave-motion in 138 
Sound waves 

in gases 
adiabatic nature of 62-4 

288 Gases, liquids and solids 

Sound waves - contd 
in gases - contd 

velocity of 60-62 
in solids 138, 174 
Specific heat 
definition of 29 
of perfect gas 35 
theory of 85-7 
of solids 171-2 
classical theory 172 
Debye theory 173-9 
Einstein theory 172-3 
Spreading, of surfactants 4 
Standing wave, equivalence of, with 

quantum oscillator 174 
States of matter 

see Gases, Liquids and Solids 
Statistical mechanics 32 
Stern 85 

Stirling approximation 150 
Stokes' law, for resistance of sphere in 

viscous medium 226 
Strain, definition of 134-5 
Streamlines, definition of 223 
Stress multipliers, role of cracks as 

Superheating 220 
Superposition of stresses 156 
Surface area, determination of, by 

heat of wetting 214 
Surface energy 
of liquids 
free 212 

molecular model of 213-14 
total 214 
of solids 
determination of 128-30 
non-identity of with line tension 

role of, in wetting 215, 222 
theory of 127-8 
Surface tension of liquids 212 
effect of surfactants on 5 
relation of, with surface energy 213 
Survival equation in gases, kinetic 

theory of 70 
Susceptibility, magnetic 
definition of 256 
theory of 256-74 

Sutherland's constant, values of 49 

concept of 23 
practical scale 23 
thermodynamic scale 31 
Temperature discontinuity at walls, role 
of, in thermal conductivity of gases 
Tensile strength 158 
Thermal capacity 
of electrons 
classical 1 88 
Fermi 190 
of gases 35,85-7 
of liquids 206 
of solids 
classical 171-2 
Debye 173-9 
Einstein 172-3 
Thermal conductivity 
of gases 55 

relation of with viscosity 57 
general mechanisms of 184-7 
of solids 187 
good conductors 188-9 
poor conductors 187 
Thermal diffusivity 186-7 
Thermal expansion 
of gases 33 
of solids 
elementary theory 132-3 
Gruneisen theory 179-84 
role of free energy in 183 
Thermal properties of solids, general 

first law 26 

relation of, with probability 32 
second law 30 
zeroth law 24 
Thixotropy, definition of 234 
Total surface energy 214 
Transport phenomena in gases 
boundary conditions 53-5, 57 
theory of, advanced 68-70 
theory of, elementary 50-60 
Troposphere, temperature drop in 40 
Turbulence 225 

289 Index 

Ultimate tensile strength, relation of, 

with hardness 158 
Unimolecular films 5 
Upper yield point, dislocation theory of 


in liquids 199 
in solids 201 
Van der Waals equation of state 
of a gas 101-4 
of a liquid 194-5 
relationship to 
Boyle temperature 107 
critical coefficient 1 14 
critical points 108-11 
values of a and b in 11 1-3 
Van der Waals forces 
comparison of, with gravitational 

forces 21 
normal 14 
retarded IS 
role of, in colloids 16 
role of, in compressibility of solids 
Van Vleck, comments of, on magnetic 

theory 269 
Vapour pressure 
effect of pressure on 211-12 
effect of temperature on 210 
molecular theory of 208 
Vibrational frequency 
in liquids 201, 230 
in solids 140, 172-9 
Vickers hardness test 157 
Virial equation of state 97 
of gases 
advanced theory 68-70 
elementary theory 50-3, 228 
role of mean-free path in 53-4 
role of slip at walls in 54-5 
of glasses 123,198 
of liquids 
effect of pressure on 232, 233 
effect of temperature on 231, 233 
elementary theory 229 
Eyring theory 230-33 

Viscosity - contd 

of liquids - contd 
role of, in lubrication 226-8 
Viscosity, non- Newtonian 234 
Viscous resistance 

Poiseuille's equation 226 

Stokes' law 225-6 
Volume strain 135 
von Laue's X-ray diffraction 6, 7 
Vortices 225 

Wainwright's molecular dynamic study 

of liquids 203 
Walls, attraction of, for gas molecules 

Wave propagation 

in gases 60-4 

in solids 139 
Waves, standing, role of, in Debye 

theory 174-9 
Wiedemann-Franz relation 190, 

Wilson cloud chamber, nucleation in 


characteristics of 154 

dislocation mechanism of 163 

X-rays, use of, in determining 
Avogadro's number 6 

Yield criteria 

for brittle failure 165 

for plastic flow 155 

summary of 170 
Yield stress 

in Bingham fluids 234 

in ductile solids 154 
Young's equation of wetting 215 
Young's modulus of elasticity 

definition of 131 

theory of 131-2,138-9 

Zartman and Ko, method of, for 
determining velocity distribution of 
metal atoms 84 

Zeroth law, of thermodynamics 24 

290 Gases, liquids and solids 

Penguin library of physical 
sciences: Physics/Chemistry 


The Penguin Library of Physical Sciences 

is a new, carefully integrated series of 
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It has been traditional to treat gases, liquids 
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molecules and the forces between them, and 
lead to a discussion of the concept of heat. 
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gases and solids, discussed in terms of 
intermolecular forces. The treatment is 
extended to the liquid state, the Cinderella of 
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Dr D. Tabor, F.R.S., is Reader in Physics in the 

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